Examples of common false beliefs in mathematics



The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.


Posted 2010-05-04T21:02:58.510

Reputation: 18 667

3Most examples are fantastic especially for those preparing for qualifying/comprehensive exams. – Unknown – 2010-11-22T19:07:53.953

1In addition to common false beliefs, I find something somewhat amusingly alleged to be a common false belief: Some time around 2003 or 2004, when Wikipedia was less developed than it later became, its article about the product rule asserted that the derivative of a product of two functions is different from what "most people think" it is. Then it said "Most people think that $(fg)' = f'g'$. – Michael Hardy – 2010-12-18T20:05:35.747

5It's almost surely time for this to be closed. Flagging for moderator attention. – Todd Trimble – 2011-10-06T12:47:27.400

4I would vote to close at this point if I didn't have superpowers. It is a great question, but perhaps 17 months is long enough. – S. Carnahan – 2011-10-06T15:48:13.770

23Meta created http://tea.mathoverflow.net/discussion/1165/examples-of-common-false-believes/ – None – 2011-10-08T14:27:20.867

Sorry for being late. Two common false beliefs: 1. Any ring epimorphism is surjective. 2. Suppose given a short exact sequence X'->X->X'' in an abelian category A. If a full subcategory B of A contains X' and X, but not X'', then X'->X does not have a cokernel in B. (Wrong for A = Z-mod, B = Z-free, (X'->X->X'') = (Z -2-> Z -> Z/2).) – Matthias Künzer – 2011-10-08T14:38:58.933

15I vote not to close – Gil Kalai – 2011-10-08T20:28:25.620

3@Matthias: the epimorphism thing might stem not so much from a false belief as from unfortunate terminology. For many people, the definition of epimorphism is surjective homomorphism. Presumably this definition predates the category-theoretic one by many decades. – Thierry Zell – 2011-10-09T20:55:34.463

@Thierry: As far as I know, "epimorphism" is Bourbaki terminology. I think Weil insisted on not mixing Greek and Latin at this point. So yes, you're right, since Bourbaki's point of view is "sets with structure", the definition via surjectivity is the original one. – Matthias Künzer – 2011-10-13T06:09:52.160

Dear @Matthias, what was the proposed mixture of Greek and Latin ? – Georges Elencwajg – 2013-10-15T14:46:13.803

@GeorgesElencwajg I think the point is that surjective homomorphism would be such a mix (the former being 'Latin' and the latter 'Greek', at least in an ethymological sense). – None – 2013-10-15T17:11:39.400

1@quid: yes, that's a possibility. I know that long ago some purists objected to television for the same reason. – Georges Elencwajg – 2013-10-15T18:33:20.353

1@Georges Elencwajg: if I recall correctly, someone suggested "unimorphism" (Latin/Greek-mixture), but Weil insisted on "monomorphism". – Matthias Künzer – 2013-11-07T08:58:24.480

This is such a wonderfull question! – CSA – 2014-06-30T17:18:39.150

2Over $200$ false beliefs so far… maybe true beliefs are even more, but certainly not as popular! – Pietro Majer – 2015-02-17T02:18:49.780

90I have to say this is proving to be one of the more useful CW big-list questions on the site... – Qiaochu Yuan – 2010-05-06T00:55:08.100

24The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. – Unknown – 2010-05-22T09:04:07.617

one typical mistake in matrix algebras: positive matrices must have positive entries. (However, for example $\begin{pmatrix} 1 & -1 \ -1 & 1 \end{pmatrix}$ is positive as well, since this matrix is self-adjoint and has non-negative eigenvalues) – Ypsilon – 2016-10-31T08:51:58.777

Students thinking that the field $\mathbb{F}_4$ is the ring $\mathbb{Z}/4\mathbb{Z}$... – Sabrina Gemsa – 2017-04-28T15:03:18.317

People think that in a complete lattice $T$, if $M\subset T$, then $\operatorname{inf} M\leq \operatorname{sup} M$ – Max – 2017-07-13T13:39:34.353

2I'm voting to close this question as off-topic because enough false beliefs already – Mikhail Katz – 2017-09-20T08:37:42.163

wow, this will soon reach 666 votes... a nice score for a question about false beliefs – Pietro Majer – 2017-10-22T12:33:21.357

20wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. – Suvrit – 2010-09-20T12:39:27.377

19It's a thought -- I might consider it. – gowers – 2010-10-04T20:13:17.327



For vector spaces, $\dim (U + V) = \dim U + \dim V - \dim (U \cap V)$, so $$ \dim(U +V + W) = \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W), $$ right?


Posted 2010-05-04T21:02:58.510

Reputation: 4 754

6Just out of interest, is there a (true!) formula for the dimension of $U+V+W$ if one knows only the dimensions which appears in the false formula above? – Mark – 2011-03-03T21:29:41.960

10@Mark: Given three distinct lines $U,V,W$ through the origin, you can compute the RHS but not the LHS. – Gerald Edgar – 2011-04-10T16:51:12.623

1Hm, true. Thanks. – Mark – 2011-04-13T09:17:30.320

Is this related to Stein's Example? – Kaveh Khodjasteh – 2011-08-24T17:10:00.080

pity one can not add the answer to favorites list) – Olga – 2013-09-27T10:17:21.190

Just last week I made this mistake in a "proof". Clearly I should check this list more often. – arsmath – 2013-12-03T10:20:05.773

3$dim(U+V+W)=d(U+V)+d(W)-d((U+V)\cap W)$ then the equality above is true iff $dim((U+V)\cap W)= dim(U\cap W)+ dim(V\cap W)- dim(U\cap V \cap W)$ i.e. iff $dim((U+V)\cap W)=dim ((U\cap W)+(V\cap W))$ iff (in finite dimention) $(U+V)\cap W=(U\cap W)+(V\cap W)$. – Buschi Sergio – 2015-01-09T14:59:40.087

4getting bad flashbacks about this one... good example, though – Yemon Choi – 2010-05-04T22:53:04.947

120Wait, that isn't true? – Simon Rose – 2010-05-04T23:19:51.080

152Take three distinct lines in R^2 as U, V, W. All intersections have 0 dimensions. The LHS is 2, the RHS is 3. The problem is that $(U+V)\cap W \neq U\cap W + V\cap W$. – Willie Wong – 2010-05-04T23:38:13.200

26Take 3 lines in $\mathbb{R}^2$... – Tom Smith – 2010-05-04T23:38:30.107

36This is perhaps a shameful comment for math overflow, but: ROFL (in the best possible sense) :-) excellent answer! – Kevin McGerty – 2010-05-05T00:26:16.573

3This is actually true for Euler Characteristic. – Harry Gindi – 2010-05-08T16:53:24.410

8@Tilman: Only a remark not related to the topic: The identity $$\dim (U + V) = \dim U + \dim V - \dim (U \cap V)$$ is valid only for finite dimensional spaces, but if one writes it as follows $$\dim (U + V) + \dim (U \cap V)= \dim U + \dim V$$ it is valid for all vector spaces. – Alireza Abdollahi – 2016-03-25T07:23:56.403

10100 upvotes! The first "Great Answer" badge! (Besides Anton's fluke from the moderator election.) – Kevin H. Lin – 2010-07-20T18:14:30.310


Everyone knows that for any two square matrices $A$ and $B$ (with coefficients in a commutative ring) that $$\operatorname{tr}(AB) = \operatorname{tr}(BA).$$

I once thought that this implied (via induction) that the trace of a product of any finite number of matrices was independent of the order they are multiplied.

Jason DeVito

Posted 2010-05-04T21:02:58.510

Reputation: 3 777

1In fact, the result applies to the eigenvalues: the eigenvalues (non-zero eigenvalues if you allow non-square matrices) are invariant under cyclic permutations. That is sometimes very useful. – kjetil b halvorsen – 2013-01-14T21:14:29.983

11Indeed. I never thought much about this before, but clearly this only implies the trace of a product is invariant under cyclic permutations.

I bet there is some fact from the representation theory of the symmetric group lurking here, but am too lazy to think about it... – Nate Eldredge – 2010-05-04T21:16:39.013

59In fact Tr$(AB)=$Tr$(BA)$ holds also for non-square matrices $A,B$ for which both $AB$ and $BA$ are defined. Now for determinants, det$(AB)$=det$(BA)$ holds for square matrices, but of course not for non-square matrices (consider the case where $A$ is a column vector and $B$ a row vector). – user2734 – 2010-05-04T21:46:02.383

28@Nate: If you want high-powered generalities, the most general situation I know where one can prove this statement is in a ribbon category. These have a graphical calculus where tr(ABC...) corresponds to a closed loop on which A, B, C... sit as labels in order, which clearly shows that the only invariance one should expect is under cyclic permutation. See, for example, the beginning of Turaev's "Quantum Invariants of Knots and 3-Manifolds." – Qiaochu Yuan – 2010-05-04T22:32:35.907

1Also, in Penrose's diagrammatic notation, composition AB is represented by a line from the top of B to the bottom of A, and the trace of A is a line from the top of A to the bottom of A. – Marcos Cossarini – 2010-05-05T00:07:17.987

@Marcos: using Penrose's diagrammatic notation for things with only two indices is a bit of an overkill. It also doesn't show that generically the only invariant we expect is from cyclic permutations, since sometimes weird tangles of lines in the diagram can be unraveled... – Willie Wong – 2010-05-05T00:23:08.830

5@Harry, if you think about what happens when you split a product $abcdefgh$ in the middle and interchange the two halfs, you'll see where Nate is going... – Mariano Suárez-Álvarez – 2010-05-05T03:30:01.730

22@unknown: nonetheless, the characteristic polynomials of AB and BA are the same up to a power of $\lambda$ (A is m by n and B is n by m), which generalizes both properties – Victor Protsak – 2010-05-05T06:54:28.900

8@Victor Protsak: Nice! BTW, one way to get what you say is from det$(I_m+AB)=$det$(I_n+BA)$, which funnily doesn't hold for the trace in case of non-square matrices (there is a difference of $m-n$). – user2734 – 2010-05-05T07:44:54.850

Yes, but losing that property is a small price for being able to say "cyclicity of the trace". – Emilio Pisanty – 2015-05-05T19:55:43.757

@QiaochuYuan could you provide an easy to access (explicit aproach) material for 'I know where one can prove this statement is in a ribbon category. These have a graphical calculus where tr(ABC...) corresponds to a closed loop on which A, B, C... sit as labels in order, which clearly shows that the only invariance one should expect is under cyclic permutation' – Turbo – 2016-01-02T20:33:00.477

@QiaochuYuan does it have anything to do with ribbon graphs? – Turbo – 2016-01-02T20:37:56.687

@NateEldredge There are some nontrivial application for the fact you mentioned9invariance of the trace under full-cyclic permutation).There is an operator theoretical proof of the Gauss Bonnet theorem in the book NCG by Alain Connes. In that proof the invariant of trace is uded. – Ali Taghavi – 2016-02-26T21:07:26.977

1If M is a matrix permuting coordinates, then $tr(M)$ is the number of fixed points of the relative permutation! – Mircea – 2010-06-06T19:41:45.490


AB and BA share the same invertible part : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=112209

– Yoo – 2010-06-15T16:59:13.377


Many students believe that 1 plus the product of the first $n$ primes is always a prime number. They have misunderstood the contradiction in Euclid's proof that there are infinitely many primes. (By the way, $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ is not prime and there are many other such examples.)

Much later edit: As pointed out elsewhere in this thread, Euclid's proof is not by contradiction; that is another widespread false belief.

Much much later edit: Euclid's proof is not not by contradiction. This is another very widespread false belief. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. In fact, if the derivation of an absurdity or the contradiction of an assumption is a proof by contradiction, then Euclid's proof is a proof by contradiction. Euclid says (Elements Book 9 Proposition 20): The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime.

Nb. The above edits were not added by the OP of this answer.

Edit on 24 July 2017: Euclid's proof was not by contradiction, but contains a small lemma in the middle of it that is proved by contradiction. The proof shows that if $S$ is any finite set of primes (not assumed to be the set of all primes) then the prime factors of $1+\prod S$ are not in $S$, so there is at least one more prime than those in $S.$ The proof that $\prod$ and $1+\prod$ have no common factors is the part that is by contradiction. All of this is shown in the following paper: M. Hardy and C. Woodgold, "Prime simplicity", Mathematical Intelligencer 31 (2009), 44–52.

Ravi Boppana

Posted 2010-05-04T21:02:58.510

Reputation: 746

16Wait, that integral comes out to $10/3-\pi$... – Harry Altman – 2010-10-20T04:18:50.727

20@Harry Altman, the integrand should be $\frac{x^4 \cdot (1 - x^4)}{1 + x^2}$. – muad – 2010-10-20T14:45:01.080

13Having thought I understood Euclid's proof, I was puzzled to learn that it fails for the ring of power series in one variable over a field. The teacher pointed out how crucial it is to check in the argument whether 1 + abcd....ef is a unit, something usually glossed over. Actually when I recalled Euclid's argument, he exclaimed "Yes, but Eucid was intelligent!" – roy smith – 2010-11-21T05:51:16.510

8I don't think I'd ever heard of the approximation 22/7 for $\pi$ until I moved to the US. Probably the people who designed the school curriculum in France decided that people would get confused over this, and I guess they were right! I've read horror stories like this lone English major sticking up for the truth in a room full of math teachers who seemed to believe that $\pi$ was rational. And anyway, you might as well use 3.14 as an approximation. – Thierry Zell – 2010-11-27T16:16:03.040

42Both Robert Bruner and muad have this wrong: The integral is $$ \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx = \frac{22}{7}-\pi. $$ – Michael Hardy – 2011-02-06T20:39:27.160

10Does anybody else remember Archimedes Plutonium (or whatever his name is now) from USENET? He was convinced that Euclid's proof was wrong and that his trivial modification was the first correct proof. This example came up in discussion, when people suggested that ironically AP's proof was wrong (although actually it was also correct). – Toby Bartels – 2011-04-04T06:39:04.293

1I was going to rollback because it would seem the edits should be comments (especially as they are not by the OP, even if this is a CW). I think the edits should probably be comments. – Benjamin Steinberg – 2013-04-16T19:40:43.067

1@RaviBoppana Actually 0.999... is not equal to 1 for the very reason that the first number is is an infinite sequence of the integer 9 and the second one is a sequence of one integer: 1. They are 2 different sequences. But in arithmetic, in the decimal model, for the sake of cohenrence, the infinite development 0.999... is identified to 1. Thus, the professor of your daughter was not really wrong, the sentence "0.999... is not equal to 1" is not precise enough (it's a trap) and contains an innuendo not familiar for people outside the strict community of self-called "mathematicians" ;-) – Patrick I-Z – 2013-12-02T23:42:03.347

12@Patrick 0.999... is not a sequence, it's a number (if you don't get this, you haven't understood decimal notation). The expression $a_0.a_1a_2a_3...$ (where the $a_i$ for $i>0$ are digits, i.e. numbers between 0 and 9) denotes the real number $a0+\sum{n\in\mathbb N}\frac{a_n}{10^n}$. It is possible to check that, if $a_0=1$ and $a_i=0$ for $i>0$, we get the same real number that we get when $a_0=0$ and $a_i=9$ for $i>0$. I fail to see how this is "a trap", it's clear to me that the teacher that we're talking about was wrong (and, being a teacher, should have known better) – David FernandezBreton – 2013-12-08T15:28:38.193

3@DavidFernandezBreton An infinite sequence defines a unique number, as you write correctly. The map $\Phi : (an){n=1}^\infty \mapsto \lim{N\to \infty} \sum{n=1}^N a_n 10^{-n}$ associates a sequence, for example 9,9,9 ad infinitum, to a number (limit of a converging sequence). Two different sequences can define the same number. Your notation 0,999... is just a way you refer to the infinite sequence of 9. Now, (999...) and (1) have just the same value by $\Phi$. Therefore, the decimal numbers are the set of infinite sequences after identification by $\Phi$. That's all I said. – Patrick I-Z – 2013-12-08T17:13:33.700

20I understand that, but I disagree about what denotes what. There are many ways of denoting sequences (for example, an infinite sequence of 9s can be denoted $(9,9,9,\ldots)$ or $\langle 9,9,9,\ldots\rangle$ or even just $9,9,9,\ldots$), but a number written in decimal expansion (as in $0.999\cdots$) denotes the number, not the sequence (that is, once you write the sequence without commas and with a decimal point, you're already referring to the number, i.e. to the image of the sequence under what you call $\Phi$). – David FernandezBreton – 2013-12-08T19:05:11.440

19Hence, $\langle 1,0,0,0,\ldots\rangle\neq\langle 0,9,9,9,\ldots\rangle$ but $1.000\cdots=0.999\cdots$. – David FernandezBreton – 2013-12-08T19:06:51.853

3@Patrick: Be careful not to conflate "number" with "numeral". Standard notations don't distinguish which is meant when you write a string of characters: you have to infer from context. (and usually, "number" is what is meant) – Hurkyl – 2014-02-02T22:47:07.333


@Hurkyl Hi there, I use the word "number" in the meaning defined by Dedekind, that is, a cut in the rational numbers. (Here for example http://www.amazon.com/Essays-Theory-Numbers-Dover-Mathematics/dp/0486210103)

– Patrick I-Z – 2014-02-04T00:11:52.847

161When I was 11 y.o. I was screamed at by a teacher and thrown out of class for pointing this out when he claimed the false belief stated (it wasn't class material, but the teacher wanted to show he was smart). I found the counterexample later at home. I didn't let the matter drop either... I knew I was right and he was wrong, and really had a major fallout with that math teacher and the school; and flunked math that year. – Daniel Moskovich – 2010-05-05T01:19:05.237

98@Daniel: Sorry to hear that. When my daughter Meena was the same age (11), her teacher asserted that 0.999... was not equal to 1. Meena supplied one or two proofs that they were equal, but her teacher would not budge. Maybe this is another example of a common false belief. – Ravi Boppana – 2010-05-05T02:59:55.023

82@Daniel: I've heard a worse story. A college instructor claimed in Number Theory class that there are only finitely many primes. When confronted by a student, her reply was: "If you think there are infinitely many, write them all down". She was on tenure track, but need I add, didn't get tenure. – Victor Protsak – 2010-05-05T05:38:11.463

8@Ravi More like an example of the fact that most schoolteachers in today's world-even at good schools,let alone the pathetic joke most mainstream grade schools are in America-don't really know math. – The Mathemagician – 2010-05-05T05:40:49.423

4@Andrew: It's an apocryphal story, so it may be a common false belief among the schoolteachers – Victor Protsak – 2010-05-05T07:13:04.227

2If you wanted to convince someone that this isn't true, wouldn't an easier example be 2357-1 = 1119? Sure, it has -1 instead of +1, but that doesn't matter, does it? – Zsbán Ambrus – 2010-05-05T10:57:17.203

2To Daniel Moskovich: our class had some serious disagreements with our biology teacher over simple probability problems, disguised as genetics. – Zsbán Ambrus – 2010-05-05T10:58:52.590

105This false belief leads to a proof of the Twin Prime conjecture: For every $n$, $(p_1 p_2 \cdots p_n -1, p_1 p_2 \cdots p_n +1)$ are twin primes, right? – David E Speyer – 2010-05-06T15:50:24.553

175Daniel, about the same age, I was asked to leave class for claiming that pi is not 22/7. The math teacher said that 3.14 is an approximation and while some people falsly believe that pi=3.14 but the true answer is 22/7. Years later an Israeli newspaper published a story about a person who can memorize the first 2000 digits of pi and the article contained the first 200 digits. A week later the newspaper published a correction: "Some of our readers pointed out that pi=22/7". Then the "corrected" (periodic) 200 digits were included. Memorizing digits of pi is a whole different matter if pi=22/7. – Gil Kalai – 2010-05-11T05:45:09.747


@DavidFernandezBreton: It's kind of late, but I think what Patrick meant is something along the lines of The Treachery of Images.

– tomasz – 2015-09-02T02:45:12.997

2Well, certainly a string of symbols is not a number, but it can stand for a number (or in general, for a mathematical object) and so it makes sense to enquire whether two different strings of symbols actually represent the same number (or mathematical object), in just the same way that it makes sense to say that "David FernandezBreton" and "the only current postdoc in logic at UofM" are actually the same person. That's the reason we use the "=" sign in Mathematics. – David FernandezBreton – 2015-10-03T22:02:07.767

11Ravi, you yourself are laboring under a false belief about Euclid's proof. As I have pointed out elsewhere on math overflow and in an article in the Mathematical Intelligencer, Euclid's proof was not by contradiction. What the students you refer to is not a misunderstanding of Euclid's proof, but a misunderstanding of a variation on Euclid's proof, which is not as good as the proof that Euclid actually wrote. – Michael Hardy – 2010-06-06T19:46:02.760

@MichaelHardy You made me curious. Can you give some links where I can find out why Euclid's proof is not by contradiction? – becko – 2017-07-24T14:53:30.880

1@becko : M. Hardy and C. Woodgold, "Prime simplicity", Mathematical Intelligencer 31 (2009), 44–52. $\qquad$ – Michael Hardy – 2017-07-24T16:03:01.770

@MichaelHardy Thank you. – becko – 2017-07-24T16:12:29.513

Can someone post a counter-example of the original assertion? I don't get it. – Carl Patenaude Poulin – 2017-10-11T02:06:01.120

1The original post already gave a counterexample. Namely 1 plus the product of the primes up to 13 is not prime. It's equal to 30,031, which factors into 59 times 509. – Ravi Boppana – 2017-10-11T03:02:17.230

50@Gil, After showing a colleague the integral $\int_0^1 \frac{(1-x)^4}{1+x^2} dx = 22/7 - \pi$ he assigned its calculation as an exercise. One student carried it out correctly up to $22/7-\pi$, then concluded $=0$. When asked, he truly believed, from his high school training, that $\pi = 22/7$. – Robert Bruner – 2010-08-26T19:47:52.550

9I had the $\pi=22/7$'' false belief at age 14 (don't know when I lost it, exactly), and I know exactly how it happened. I had long known that $\pi$ is about $3.14$, of course, and then in my math textbook I encountered a phrase along the linesusing $\pi=22/7$, this expression simplifies to...''. I remember the satisfaction at finally learning what ``aproximately equal'' meant. It's unclear if this was a false belief successfully reproducing, or if it was spontaneous generation. – Kevin O'Bryant – 2010-10-14T15:42:31.830


The closure of the open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space.

In a somewhat related spirit: the boundary of a subset of (say) Euclidean space has empty interior, and furthermore has Lebesgue measure zero. (This false belief is closely related to Gowers' example of the belief that there are no non-trivial open dense sets.)

More generally, point set topology and measure theory abound with all sorts of false beliefs that only tend to be expunged once one plays with the canonical counterexamples (Cantor sets, bullet-riddled squares, space-filling curves, the long line, $\sin\left(\dfrac{1}{x}\right)$ and its variants, etc.).

Terry Tao

Posted 2010-05-04T21:02:58.510

Reputation: 53 985

42Peter: actually the simplest counterexample is the open/closed ball of radius $0$, empty set vs a singleton. – Pietro Majer – 2011-08-01T15:43:11.163

@Pietro: Nice :) – Peter Samuelson – 2011-10-06T18:01:58.823

(True statement) A subset of $\mathbb{R}^n$ is Peano-Jordan measurable if and only if its boundary is Peano-Jordan measurable with measure zero. – nullUser – 2013-07-08T14:42:07.520

4In response to Qiaochu's comment, I'm surprised nobody ever mentioned that a canonical counterexample to the first claim is given by the $p$-adics: there every ball is clopen, and if the "closed radius" of the ball is $p^{-n}$, the "open radius" is $p^{-n+1}$. This is because the image of the distance function is discrete (except at distance 0). – Tim Campion – 2015-09-06T03:02:34.727

8I remember being assigned as an exercise to find a counterexample to the first statement, but I can't remember where. Rudin? – Qiaochu Yuan – 2010-06-06T23:39:08.530

19What about a space with 2 points a distance 1 apart, and the open/closed ball having radius 1? I don't remember seeing this before, though. – Peter Samuelson – 2010-06-06T23:53:29.433

6@Terry Really good examples. We can count on you to do anything but waste our time with a post, Terry. I hope you keep finding the time to post here and lend your support! – The Mathemagician – 2010-06-07T00:11:19.577

4These seem to be more "interesting mistakes" than "false beliefs", especially the last part. – Victor Protsak – 2010-06-10T06:02:48.117

"The closure of the open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space".It seems to me that this ought to be true. Since it's not, I'm led to ponder whether we have the right definitions of "metric", or "closure". How we can we develop any intuition in a situation where things that seem "obviously true" are actually false. – bubba – 2016-06-05T06:35:23.773

@bubba: at most you may complain about the terminology, certainly not about the axioms of metric spaces, and it would be foolish changing them in order to make that statement true (besides, closed ball and closure of the ball are well distinct expressions, so there is no ambiguity). – Pietro Majer – 2017-02-01T11:31:24.377

it is interesting that although this seems plausible at first sight, after being told it is false, it takes only a moment to think of a counterexample. I guess the key is that we tend to assume wrongly that an open ball is non empty, or even has lots of points in every direction. – roy smith – 2017-07-24T03:04:19.090

@QiaochuYuan Erwin Kreyzig's Introductory Functional Analysis with Applications does that, and the intended counterexample there is a discrete space. Later on he also notes that the statement is true in normed spaces. – N Unnikrishnan – 2017-07-24T08:49:49.507

@PietroMajer Most elementary textbooks usually define balls to be of radius $r>0$. – N Unnikrishnan – 2017-07-24T09:02:26.467


Here's my list of false beliefs ;-):

  • If $U$ is a subspace of a Banach space $V$, then $U$ is a direct summand of $V$.
  • If $M/L, L/K$ are normal field extensions, then the same is true for $M/K$.
  • Submodules/groups/algebras of finitely generated modules/groups/algebras are finitely generated.
  • The Krull dimension of a subring is at most the Krull dimension of the ring.
  • The Krull dimension of a noetherian domain is finite.
  • If $A \otimes B = 0$, then either $A=0$ or $B=0$.
  • If $f$ is a smooth function with $df=0$, then $f$ is constant.
  • If $X,Y$ are sets such that $P(X), P(Y)$ are equipotent, then $X,Y$ are equipotent.
  • Every short exact sequence of the form $0 \to A \to A \oplus B \to B \to 0$ splits.
  • $R[[x,y]] = R[[x]][[y]]$ as topological rings.
  • $R[x]^* = R^*$, even if $R$ is not a domain.
  • Every presheaf on a site has an associated sheaf. (Hint: the index category of the usual colimit has to be essentially small!)
  • (Co)limits may be computed in full subcategories. For example, $Spec(\prod_i R_i) = \coprod_i Spec(R_i)$ as schemes because $Spec$ is an antiequivalence.
  • Every finite CW-complex is compact, thus every CW-complex is locally compact.
  • The smash product of pointed spaces is associative (this is even false for CW complexes when you don't use the compactly-generated product!), products commute with quotients, and so on: Topologists often assume that everything behaves well, but sometimes it does not.

Martin Brandenburg

Posted 2010-05-04T21:02:58.510

Reputation: 31 905

1I would like to know more about $\mathcal{P}(X)$ equipotent to $\mathcal{P}(Y)$ not implying $X$ and $Y$ being equipotent. Is there no proof with the axiom of choice? It seems the gen. continuum hypothesis should imply it. Can you point me to some reference? – Olivier Bégassat – 2011-05-03T15:44:34.403

6Generalised continuum hypotesis implies this statement, while Martin Axiom + negation of continuum hypothesis provide a counterexample $P(\aleph_1)=P(\aleph_0)$ hence this misbelief is in fact independent of ZFC – Ostap Chervak – 2011-06-29T11:47:00.337

Olivier, you might want to check out Easton's theorem in forcing: http://en.wikipedia.org/wiki/Easton%27s_theorem

– Todd Trimble – 2012-01-28T16:40:42.367

7Amazingly enough, the splitting belief IS true if you add the innocuous-looking condition that $A$ and $B$ are finitely generated modules over a commutative Noetherian ring. (Theorem 1 from T. Miyata, Note on direct summands of modules, J. Math. Kyoto Univ. 7 (1967) 65-69) – Neil Epstein – 2012-04-26T09:42:22.517

4@Harry: Very late addendum, but for an explanation of why sheafifying over large sites really is problematic even when you assume universes, see Waterhouse (1975) - Basically bounded functors and flat sheaves. The point is that the result of sheafification depends on the choice of universe when you use universes to construct them, i.e. it is no longer intrinsic. – Zhen Lin – 2012-11-26T11:10:58.010

1Hey, I currently share a half of these allegedly false beliefs! – Michael – 2013-09-18T19:05:41.233

As a positive result, if $0\to A\to A\oplus B\to B\to 0$ is an exact sequence of finitely generated modules over a commutative Noetherian ring, then the exact sequence does split. – Mohan – 2014-01-02T16:31:42.823

5I think the "topologists assume" sentence in the last bullet is unfair; it implies topologists are making mistakes. Certainly competent topologists are not making such rookie mistakes, and are well aware of the standard counterexamples. – Todd Trimble – 2014-10-16T13:29:54.990

Wait a sec, what would be a counterexample for "The Krull dimension of a subring is at most the Krull dimension of the ring"? – Michael – 2014-10-16T18:28:39.587

2@Michael: $\mathbb{Z} \subseteq \mathbb{Q}$ – Martin Brandenburg – 2014-10-16T20:14:43.353

@MartinBrandenburg: Oh! I was thinking in terms of function rings of algebraic varieties. – Michael – 2014-10-16T20:29:21.353

86+1: you had me at "Here's my list of false beliefs". – Pete L. Clark – 2010-05-05T23:41:39.650

5I'm sure you'll have me kicking myself in a moment... but how does a short exact sequence of the form 0 --> A --> A + B --> B --> 0 fail to split? In any Abelian (indeed, additive is enough) category, since A + B is a biproduct, there's a paired map (0,1_B): B --> A + B, and a copaired map [1_A,0]: A + B --> A, which split each half of the sequence... don't they? Or were you thinking of a context for this example that's wider than Abelian categories? – Peter LeFanu Lumsdaine – 2010-05-06T15:48:04.610

16$A \to A \oplus B$ does not have to be the inclusion; likewise $A \oplus B$ does not have to be the projection. Thus the error here is: Two chain complexes, which are isomorphic "pointwise", don't have to be isomorphic. This occurs sometimes. – Martin Brandenburg – 2010-05-06T16:12:23.360

8I once made this very mistake, and it invalidates one of the main theorems of a published article I once quoted. A good reason in my opinion to specify what are the arrows when writing a sequence or a diagram: they are usually what you think they are, but hey, let's check. – Olivier – 2010-05-07T11:11:48.170

5Ooh, very nice --- a classic "check your implicit assumptions" example. Good point! – Peter LeFanu Lumsdaine – 2010-05-08T02:49:22.777

Is the one on Spec false or true-but-not-because-of-the-obvious-thing? – babubba – 2010-05-09T19:09:02.700

6The left side is quasi compact, but the right side only when $R_i = 0$ for almost all $i$. The difference can be made precise if $R_i$ are fields. Then $Spec(\prod_i R_i)$ is the Stone-Cech-compactification of the discrete space $\coprod_i Spec(R_i)$. – Martin Brandenburg – 2010-05-09T20:41:05.857

47Your fifth example reminds me of an even more plausible false belief I once held: if $A \otimes A = 0$, then $A = 0$. – Reid Barton – 2010-05-11T02:12:55.800

6@Reid Barton: Could you please provide a counterexample? – Regenbogen – 2010-05-15T10:17:36.903

24@Regenbogen: Take the abelian group $\mathbb{Q}/\mathbb{Z}$. – Steve D – 2010-05-15T13:44:16.180

5The point about presheaves and associated sheaves is one of those unimportant size issues that can be rectified by using universes and is a technical point that depends on a specific choice of set-theoretic formalism (For this reason, I suspect that Grothendieck ignores this issue in SGA4). I don't know if it really warrants inclusion on this list, since the rest of the list is so good. – Harry Gindi – 2010-05-16T06:55:37.217

3@Harry: No, it's a real problem because often you want to stay in the fixed universe; otherwise mathematics becomes pathological. For example, you could claim that every continuous functor has a left adjoint since the solution set is satisfied if we make the universe large enough for the solution set condition. But then we are not talking anymore about the same categories and functors between them! – Martin Brandenburg – 2010-05-16T15:18:50.450

3@Martin: No, that doesn't really matter as long as we keep track of relative size differences. There's no pathology there. – Harry Gindi – 2010-05-16T17:41:05.827

10I may be stupid, but what is a non-constant smooth function with df = 0 everywhere? – Lennart Meier – 2010-05-19T08:07:21.723

21$f$ is just locally constant ;-) – Martin Brandenburg – 2010-05-19T08:39:55.693

@Martin: the statement "The Krull dimension of a noetherian domain is finite." is my false belief today :) . doesn't this implied by KRULL'S PRINCIPAL IDEAL THEOREM? I mean if $R$ be noetherian ring, height of every maximal ideal is finite. and $\dim R$ is $sup$ of these heights. – user 1 – 2016-02-04T08:03:38.710

@mohan do you have a reference? isnt this contradiction with Martin's statement? – user 1 – 2016-02-04T08:18:58.750

@user1 I do not have a reference, but it was mentioned with reference here earlier by others (e. g. Graham Leuschke) too. The proof, while not trivial, can be worked out and I would be happy to post one somewhere (how?) if you so desire. – Mohan – 2016-02-04T14:34:59.857

1I saw doctorate thesis defence when one of reviewers, prominent one, claimed that: "If f is a smooth function with df=0 , then f is constant." is true, and then work has serious flaw. f in this work was topological invariant with df=0 but clearly there was different topological charges here, not only one ( and it was shown in the work).... – kakaz – 2010-06-10T10:44:17.407

@user1 but there could be maximal ideals of many different heights! – Will Sawin – 2017-01-14T03:19:46.447

@Will Sawin yes, thank you. – user 1 – 2017-01-14T13:22:01.763

Could you give an example for two sets whose power sets are equipotent whereas they are not equipotent? – Fawzy Hegab – 2017-03-08T21:12:42.243

@FawzyHegab: It depends on the choice of the model of set theory whether this is true or not. – Martin Brandenburg – 2017-04-12T11:16:46.990

The fact that the inclusion $\mathbb{Z} \subset \mathbb{Q}$ does not preserve dimension can be expressed by saying that $\mathbb{Q}$ is zero-dimensional, but not hereditarily zero dimensional. These are studied in the book Zero-Dimensional Commutative Rings, edited by David Dobbs. – Somatic Custard – 2017-12-21T16:19:34.260

@MartinBrandenburg: I don't understand the locally-constant hint. Would you mind giving an actual counterexample? – Mehrdad – 2017-12-28T09:10:32.477


I don't know if this is common or not, but I spent a very long time believing that a group $G$ with a normal subgroup $N$ is always a semidirect product of $N$ and $G/N$. I don't think I was ever shown an example in a class where this isn't true.

Qiaochu Yuan

Posted 2010-05-04T21:02:58.510

Reputation: 73 568

2maybe the easiest way to see this, as suggested by Kevin's example, is to think of abelian groups and ask whether every subgroup is a direct factor. I.e. this has little to do with true semi direct products, and more to do, as Fabrizio observed, with splitting maps. – roy smith – 2011-04-14T17:57:14.360

3I tripped up on this one for a VERY long time too! Given experiences cited here, I would strengthen Kevin's comment and say it proves :) we do a terrible job explaining semidirect products. And I second the comment about short exact sequences. – WetSavannaAnimal aka Rod Vance – 2011-05-04T00:35:36.307

6Argh! Me too! What is a good example? – Nate Eldredge – 2010-05-04T21:28:17.857

136umm Z/4Z contains Z/2Z? – Kevin Buzzard – 2010-05-04T21:30:51.943

3It is a sad state of things, but my impression is that most people coming out of the standard introductory course to groups have more or less the sam belief :( – Mariano Suárez-Álvarez – 2010-05-04T23:33:43.653

63This suggests that we do a terrible job of talking about semi-direct products no? – Kevin McGerty – 2010-05-05T00:27:35.603

49Schur--Zassenhaus says that this is true if $N$ and $G/N$ have coprime orders, so there is some intrinsic pressure in the subject towards this. Coupled with the fact that it is true for the first non-trivial non-abelian example ($A_3$ inside $S_3$), it's easy to see how this misconception arises. – Emerton – 2010-05-05T01:49:03.157

19Remember being confused by this too. It became much clearer when I formally was taught about short exact sequences. Then you can see exactly the obstruction to such a decomposition. – Fabrizio Polo – 2010-05-10T11:14:21.663

It took me a long time to realize that was false as well... Still being an undergrad, I often catch myself trying to use that "theorem". – Cory Knapp – 2010-06-09T01:42:53.143

This is false, but something very nice is true and related to this: $F : G \text{-Grp} \rightarrow G \uparrow \text{Grp}$ where $\phi : G \rightarrow \text{Aut}(H)$ is sent to $G \rightarrow H \rtimes_{\phi} G$ where $g \mapsto (1, g)$ is a left adjoint. So the exact sequences $0 \rightarrow N \rightarrow G \rightarrow G / N \rightarrow 0$ which come from semi-direct products are the "free" ones in a sense. – Dean Young – 2018-02-04T04:33:28.603

Or another condition for an exact sequence $0 \rightarrow N \rightarrow G \rightarrow H \rightarrow 0$ to be isomorphic to one of the form $0 \rightarrow N \rightarrow N \rtimes H \rightarrow H \rightarrow 0$ is if $G \rightarrow H$ has a section. – Dean Young – 2018-02-04T09:53:03.987


These are actually metamathematical (false) beliefs that many intelligent people have while they are learning mathematics, but usually abandon when their mistake is pointed out, and I am almost certain to draw fire for saying it from those who haven't, together with the reasons for them:

The results must be stated in complete and utter generality.

Easy examples are left as an exercise to the reader.

It is more important to be correct than to be understood.

(Applicable to talks as well as papers.)

Reasons: 1. Von Neumann is in the audience. 2. This is just a generalization of Lemma 1.2.3 in volume X of Bourbaki. 3. The results are impressive and speak for themselves.

Victor Protsak

Posted 2010-05-04T21:02:58.510

Reputation: 11 982

16IMO "It is more important to be correct than to be understood" is not a false belief. – Michael – 2014-10-16T18:32:46.397

24I definitely agree with the OP that "It is more important to be correct than to be understood" is false - in the context of giving mathematical talks. Or perhaps, it's fairer to say that being understood is more important than being 100% correct. Talks are about the listener, not about the speaker. – Greg Martin – 2015-03-05T20:15:06.493

12@GregMartin: When you are giving a talk, sure. When you are giving a lecture, maybe, but you should give an indication of where you are imprecise. When you are writing a paper, most definitely not. – tomasz – 2015-09-02T03:00:04.410

2@Michael: Victor and you are both right: you should be correct $\mathbf{and}$ understandable. So that the audience can understand, that you are correct. – co.sine – 2017-05-29T05:17:37.127

I'm not I like the usage of 'metamathematical' here because the word can have a precise and formal meaning. I can't think of anything else credible, though. Somewhere between mathematical and pedagogical? – James Smith – 2017-08-11T13:49:36.533


a student, this afternoon: "this set is open, hence it is not closed: this is why [...]"


Posted 2010-05-04T21:02:58.510

Reputation: 1 378

7I think we need more detail to be fair. Was this afternoon's student perhaps contemplating a non empty proper subset of R? – roy smith – 2011-04-14T19:24:19.740


See also http://www.arsmathematica.net/archives/2012/01/20/hitler-on-topology/

– Todd Trimble – 2012-01-28T16:45:24.487

95The terminology is rather unfortunate. – Nate Eldredge – 2010-05-05T14:39:20.630

1Yikes,that student needs a sit-down about the facts of life in topology. – The Mathemagician – 2010-05-05T16:56:39.207

56Either that or topologists need a sit-down about the facts of life in life, where they are told how unfortunate their notation is... – Kevin Buzzard – 2010-05-05T20:25:51.597

151Munkres is fond of saying "sets are not doors." – Qiaochu Yuan – 2010-05-05T20:53:37.927

3On the other hand, one can say "open the door" and "close the door" in reference to a door that is slightly ajar. – CrazyHorse – 2010-05-05T21:52:39.613

27So are you saying sets are closed, open, clopen, or ajar? ;) – jeremy – 2010-05-06T02:17:00.820

9some students are a rich source of false beliefs. Try asking whether the product of two odd functions on R is odd or even. – Pietro Majer – 2010-05-21T15:23:32.193

60On my office door I once put "clopen the door" – hypercube – 2010-05-26T22:52:48.790

3I like that "Sets are not doors", I can say that I have thought too fast and made this assumption and ended up proving something that couldn't possibly be true >< – Michael Hoffman – 2010-05-31T08:03:50.247

6When is a set not a set? – jd.r – 2010-06-05T22:47:10.143

86Actually, topologists have studied spaces where every set is open or closed (or both, of course), and they're called "Door spaces".... – Henno Brandsma – 2010-06-06T11:46:37.717

17The mere existence of the adjective "half-open", as in "the half-open interval [1,2)", is a fairly good antidote to this, even if the notion of half-openness per se does not extend particularly well beyond the interval case. – Terry Tao – 2010-06-07T18:11:50.657

1@NateEldredge Sorry, I am just an undergrad student reading this out of interest. I want to make sure I know where the mistake is here: An open set is $(2, 4)$, a closed set is $[2, 4]$, but the student failed to take into account sets such as $(2, 4]$ and $[2, 4)$, which are neither open nor closed? – Ovi – 2016-08-19T15:06:56.043

@Ovi: Yes, that's right. – Nate Eldredge – 2016-08-19T15:16:37.973

7@Ovi No, that is not right. The student said: The set is open, hence not closed." This is wrong because there are sets which are open and closed, not because there are sets that are neither. For instance, in $\Bbb R$ (equipped with its standard topology), the sets $\Bbb R$ and $\varnothing$ (the second is the empty set) are both open and closed. In fact, they are the only open and closed sets in $\Bbb R$, since $\Bbb R$ is connected. – Danu – 2016-12-07T01:05:48.200

1To all the people who find fault with topologists' terminology, sets should be compared with rooms, not doors in the first place, should they not? And the room analogy fits this bill well - rooms can be open, closed, partially open or partially closed to any degree. – N Unnikrishnan – 2017-07-24T10:32:38.083

2The terminology is poor, let it be doors or rooms or whatever. It's common sense that objects which can be open or close, usually are in only one of these states: doors, chests, safes, lockers, etc. These words are considered opposites! It's like define some sets to be hot and cold and then say there are some sets which are hot and cold at the same time. Please, if this is the case, just don't use these words. Good notation and good terminology are important. – Integral – 2017-08-08T02:48:32.747


Here are two things that I have mistakenly believed at various points in my "adult mathematical life":

For a field $k$, we have an equality of formal Laurent series fields $k((x,y)) = k((x))((y))$.

Note that the first one is the fraction field of the formal power series ring $k[[x,y]]$. For instance, for a sequence $\{a_n\}$ of elements of $k$, $\sum_{n=1}^{\infty} a_n x^{-n} y^n$ lies in the second field but not necessarily in the first. [Originally I had $a_n = 1$ for all $n$; quite a while after my original post, AS pointed out that that this actually does lie in the smaller field!]

I think this is a plausible mistaken belief, since e.g. the analogous statements for polynomial rings, fields of rational functions and rings of formal power series are true and very frequently used. No one ever warned me that formal Laurent series behave differently!

[Added later: I just found the following passage on p. 149 of Lam's Introduction to Quadratic Forms over Fields: "...bigger field $\mathbb{R}((x))((y))$. (This is an iterated Laurent series field, not to be confused with $\mathbb{R}((x,y))$, which is usually taken to mean the quotient field of the power series ring $\mathbb{R}[[x,y]]$.)" If only all math books were written by T.-Y. Lam...]

Note that, even more than KConrad's example of $\mathbb{Q}_p^{\operatorname{unr}}$ versus the fraction field of the Witt vector ring $W(\overline{\mathbb{F}_p})$, conflating these two fields is very likely to screw you up, since they are in fact very different (and, in particular, not elementarily equivalent). For instance, the field $\mathbb{C}((x))((y))$ has absolute Galois group isomorphic to $\hat{\mathbb{Z}}^2$ -- hence every finite extension is abelian -- whereas the field $\mathbb{C}((x,y))$ is Hilbertian so has e.g. finite Galois extensions with Galois group $S_n$ for all $n$ (and conjecturally provably every finite group arises as a Galois group!). In my early work on the period-index problem I actually reached a contradiction via this mistake and remained there for several days until Cathy O'Neil set me straight.

Every finite index subgroup of a profinite group is open.

This I believed as a postdoc, even while explicitly contemplating what is probably the easiest counterexample, the "Bernoulli group" $\mathbb{B} = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$. Indeed, note that there are uncountably many index $2$ subgroups -- because they correspond to elements of the dual space of $\mathbb{B}$ viewed as a $\mathbb{F}_2$-vector space, whereas an open subgroup has to project surjectively onto all but finitely many factors, so there are certainly only countably many such (of any and all indices). Thanks to Hugo Chapdelaine for setting me straight, patiently and persistently. It took me a while to get it.

Again, I blame the standard expositions for not being more explicit about this. If you are a serious student of profinite groups, you will know that the property that every finite index subgroup is open is a very important one, called strongly complete and that recently it was proven that each topologically finitely generated profinite group is strongly complete. (This also comes up as a distinction between the two different kinds of "profinite completion": in the category of groups, or in the category of topological groups.)

Moreover, this point is usually sloughed over in discussions of local class field theory, in which they make a point of the theorem that every finite index open subgroup of $K^{\times}$ is the image of the norm of a finite abelian extension, but the obvious question of whether this includes every finite index subgroup is typically not addressed. In fact the answer is "yes" in characteristic zero (indeed $p$-adic fields have topologically finitely generated absolute Galois groups) and "no" in positive characteristic (indeed Laurent series fields do not, not that they usually tell you that either). I want to single out J. Milne's class field theory notes for being very clear and informative on this point. It is certainly the exception here.

Pete L. Clark

Posted 2010-05-04T21:02:58.510

Reputation: 49 252

In 1st first ex. it is aparent when you view k a local field and look $k((x))((y))$ as a $3$-local field. Then $k((x))((y))=k((y)){{x}}$, where the field $k{{x}}$ is defined as the set of Laurent series $f$ of $y$ with coeff. in $k$ such that the $k$ valuation of the coeeficients of $f$ is bounded below and the coefficients tends to zero as the exponent tends to $-\infty$. As clear from the definition $k((x)){{y}}$ is not isomorphic to $k((y)){{x}}$ hence both can not isomorphic to $k((x,y))$. for more details you can check http://msp.warwick.ac.uk/gtm/2000/03/gtm-2000-03p.pdf

– safak – 2010-12-02T01:52:59.827

6Milne also, in his notes on field and Galois theory, takes the time to point out (and prove using Zorn's lemma and the group $\mathbb{B}$ above) that the absolute Galois group of $\mathbb{Q}$ has non-open subgroups of index $2^n$ for all $n>1$. He adds as a footnote a quote of Swinnerton-Dyer where he mentions the "unsolved [problem]" of determining whether every finite index subgroup of $G_\mathbb{Q}$ is open or not, observing that this problem seems "very difficult." – Keenan Kidwell – 2010-05-05T12:45:53.560

3Nice examples! Actually it is known that any finite groups arises as a Galois group over K=C((x,y)). Since K is Hilbertian, it is enough to prove it for K(t). Now, we know that if L is a large field (ie any smooth L-curve has infinitely many L-points as soon as it has one), then any finite groups arises as a Galois group over L(t) (see F.Pop, Embedding problems over large fields, Ann. of Math., 1996). And F. Pop recently proved that if R is a domain which is complete wrt a non-zero ideal (Henselian's enough), then its fraction field is large (see Henselian implies Large on his webpage). – Jérôme Poineau – 2010-05-05T13:52:27.173

@JP: Thanks very much for the information. I was just thinking that this should be a case close to the border of the IGP and that I should check up on what is known. – Pete L. Clark – 2010-05-05T17:49:46.593

21@Pete: I remember once reading a paper of Katz and being bewildered by what he was saying until I realised that Q_p[[x]] was much bigger than Zp[[x]] tensor{Z_p} Q_p. – Kevin Buzzard – 2010-05-05T20:19:26.807

I like that one! – Paul Balmer – 2010-05-08T18:45:22.357

Kevin: I'm quite fond of this distinction myself. It's why $\mathbb{Q}_p[[x]]$, which is the $\mathbb{Q}_p$-pro-unipotent completion of $\mathbb{Z}$, almost never comes up in Iwasawa theory. There, you're far more likely to see the small algebra. – Minhyong Kim – 2010-05-13T09:26:59.803

9It's funny that you are illustrating yourself how tricky the distinction between $k((x))((y))$ and $k((x,y))$ can be, by giving a wrong example: in fact $\sum{i \geq 0}x^{−i}y^i \in k((x,y))$. (Isn't it just $x/(x - y)$? Think a bit about convergence issues.) But I believe that $\sum{i \geq 0} x^{-i^2} y^i \not\in k((x,y))$ - and I think I can prove this using the Weierstrass preparation theorem for Laurent series over complete DVRs, or something like that. – Wanderer – 2010-07-08T18:07:39.343

@AS: Good point! I'm not sure how I missed your comment the first time around. I "fixed" my example by making it more wishy-washy. I would be very interested in seeing an explicit element in the larger field but not the smaller field, with proof. If I ask this as an MO question, would you answer it? – Pete L. Clark – 2010-07-28T07:52:57.973

Yes, I would. Go ahead! – Wanderer – 2010-07-28T13:56:13.643


Some false beliefs in linear algebra:

  • If two operators or matrices $A$, $B$ commute, then they are simultaneously diagonalisable. (Of course, this overlooks the obvious necessary condition that each of $A$, $B$ must first be individually diagonalisable. Part of the problem is that this is not an issue in the Hermitian case, which is usually the case one is most frequently exposed to.)

  • The operator norm of a matrix is the same as the magnitude of the most extreme eigenvalue. (Again, true in the Hermitian or normal case, but in the general case one has to either replace "operator norm" with "spectral radius", or else replace "eigenvalue" with "singular value".)

  • The singular values of a matrix are the absolute values of the eigenvalues of the matrix. (Closely related to the previous false belief.)

  • If a matrix has distinct eigenvalues, then one can find an orthonormal eigenbasis. (The orthonormality is only possible when the matrix is, well, normal.)

  • A matrix is diagonalisable if and only if it has distinct eigenvalues. (Only the "if" part is true. The identity matrix and zero matrix are blatant counterexamples, but this false belief is remarkably persistent nonetheless.)

  • If $\mathcal L: X \to Y$ is a bounded linear transformation that is surjective (i.e. $\mathcal Lu=f$ is always solvable for any data $f$ in $Y$), and $X$ and $Y$ are Banach spaces then it has a bounded linear right inverse. (This is subtle. Zorn's lemma gives a linear right inverse; the open mapping theorem gives a bounded right inverse. But getting a right inverse that is simultaneously bounded and linear is not always possible!)

Terry Tao

Posted 2010-05-04T21:02:58.510

Reputation: 53 985

I was puzzled by this one: "A matrix is diagonalisable if and only if it has distinct eigenvalues." until I realized you meant distinct roots of the characteristic polynomial rather than the minimal polynomial. – roy smith – 2015-09-06T02:15:14.303

30Wow. I believed that second one until now. Which is ridiculous, of course, since the operator norm of a nilpotent matrix can't be zero or else it wouldn't be a norm! – Qiaochu Yuan – 2010-06-06T23:45:38.767

2The parethentical comment in 2nd bulleted point is worded as if, $\textit{in general},$ the operator norm were equal both to the spectral radius and the largest singular value (or, perhaps, that $|A|=\rho(A)$ and $\lambda_1(A)=s_1(A).$) But for a nilpotent matrix the spectral radius is 0, whereas the operator norm and the largest singular values aren't. – Victor Protsak – 2010-06-10T07:59:18.873

2Fair enough; I've reworded the parenthetical. – Terry Tao – 2010-06-10T16:35:22.967

I guess in the last bullet point you mean a right inverse? It's easy enough to give a surjective bounded linear transformation which isn't bijective. – Mark Meckes – 2010-06-10T17:39:47.397

Also (pardon the pickiness) you presumably mean the OMT gives a continuous right inverse. – Mark Meckes – 2010-06-10T17:42:14.437

4Yes, I meant right inverse, thanks.

Getting a continuous right-inverse is actually a subtle question - the OMT only gets boundedness, which is not equivalent to continuity when one is not linear. I believe that the existence of a continuous right inverse may follow from a classical theorem of Bartle and Graves, but this is nontrivial. – Terry Tao – 2010-06-10T19:06:18.297

(I should also point out that by "bounded" I mean "maps bounded sets to bounded sets", not "maps the entire space to a bounded set".) – Terry Tao – 2010-06-10T19:08:57.310

Got it, thanks; I was thinking of applying OMT the wrong way around. – Mark Meckes – 2010-06-11T19:45:36.623

2In the last clause, there is always a continuous (usually non-linear, of course) left inverse. This is the Bartle-Graves theorem. – Tomek Kania – 2017-05-23T09:57:38.567


I once thought that if $A$, $B$, $C$, and $D$ were $n$-by-$n$ matrices, then the determinant of the block matrix $\pmatrix{A & B \\\ C & D}$ would be $\det(A) \det(D) - \det(B) \det(C)$.

Ravi Boppana

Posted 2010-05-04T21:02:58.510

Reputation: 746

3A general rule is that a formula that does not make sense in some situations must be wrong. Here, just be imaginative enough, the matrices $A$ and $D$ could be square of different sizes $p$ and $q$. Then $B$ and $C$ are not square ... Then the formula $\det A\det D−\det(BC)$ would make sense, but the latter term $\det(BC)$ is zero if $p > q$ ! – Denis Serre – 2010-10-20T10:59:04.417

There is, however, some more complicated formula similar to this, which goes like this. You want to compute the determinant of the square block matrix $ \pmatrix{A\B} $, where A has k rows. To compute it, you take all square submatrices formed by any k columns of A and multiply its determinant with the determinant of the submatrix formed by the remaining columns of B, then sum all these products with some signs to get the right answer. – Zsbán Ambrus – 2010-11-27T19:47:01.630

14@senti_today: It is fascinating how the Amer. Math. Monthly spreads and reproves lemmata from Bourbaki as theorems since they are "not widely known". The whole point of this paper is lemma 1 in Algebra ch. III §9 no. 4. – Torsten Schoeneberg – 2013-11-19T13:15:02.807


@AndrewL The belief that heavier items do not fall faster than lighter ones is false because the heavier item would exert a greater gravitational force than the lighter one. There's a Physics.SE question about it here.

– Chuu – 2014-07-15T22:07:57.403

1The result is correct when either of the four blocks is zero. – Jay – 2015-11-05T18:24:44.273

37This mistake's like assuming heavier objects fall faster then light ones. It's perfectly reasonable-just happens to be DEAD WRONG. – The Mathemagician – 2010-06-07T00:08:31.653

Hm, I think I believed that. What's a nice counterexample? – Nate Eldredge – 2010-06-07T02:19:22.277

45Make each block a $2 \times 2$ matrix with exactly one 1, in a different position in each block. You can arrange the blocks so that the total matrix is a permutation, but each block has zero determinant. – Matt Noonan – 2010-06-07T04:11:42.593

19...Also! The false formula has the wrong symmetry properties. You can swap the left and right blocks by passing the $n$ right columns over the $n$ left columns one at a time for a total of $n^2$ swaps, which affects the determinant by a factor of $(-1)^{n^2}$ but affects the proposed formula by a factor of $-1$. So at least if $n$ is even, the formula can't be right. – Matt Noonan – 2010-06-07T04:53:39.187

1I recall learning once that the formula is correct if certain pairs of the matrices commute (and maybe the word "transpose" should be in there somewhere), but I can't quite recall what the exact condition is. – Qiaochu Yuan – 2010-06-08T14:12:17.723


It seems that the answer $det(AD-BC)$ is the correct one provided $A,B,C,D$ commute pairwise. A more general result (for block matrices consisting of $k^2$ square blocks) appears in


– senti_today – 2010-06-10T18:49:13.903

1A quick reason to be false: counting the terms in the expansions, the formula would imply $(2n)!=2(n!)^2$ that is ${2n \choose n}=2$ (which is indeed true for $n=1$). – Pietro Majer – 2017-05-02T18:05:47.863


In http://en.wikipedia.org/wiki/Determinant under the "block matrices" section they give some formulas in special cases. For example, if A is invertible, then the determinant of the block matrix is $det(A)det(D-CA^{-1}B)$.

– Peter Samuelson – 2010-07-01T21:01:10.527


"Any subspace of a separable topological space is separable, too." Sounds natural.


Posted 2010-05-04T21:02:58.510

Reputation: 2 612

@MarcosCossarini: your comment leads to a question: on which side of the discontinuities between metric and topological spaces would Finsler spaces fall? – Michael – 2015-01-30T23:31:46.940

A Finsler metric (symmetric or not) can always be bounded between two Riemannian metrics, and induces the same topology that the manifold already has before considering any metric. This topology is metrisable. – Marcos Cossarini – 2015-02-02T02:23:20.233

12(and it is true of metric spaces, and natural generalizations...) – Mariano Suárez-Álvarez – 2010-05-04T23:31:04.747


This seems to be the fault of the "divorce" of second countability and separability in general topological spaces: they coincide for metrizable spaces, but for general spaces the favorable properties were split up in a custody hearing: see http://en.wikipedia.org/wiki/Separable_space#Separability_versus_second_countability. (I think I wrote this part of the article.)

– Pete L. Clark – 2010-05-05T02:14:47.400

15I think I kind of believed that until today. – Olivier – 2010-05-05T08:17:03.747

32By an amusing coincidence, I came across this for the first time a couple of days ago. There was a Cambridge exam question in 2008 where you had to show that products and subspaces of separable metric spaces were separable, and then you were given a topological space and asked to show that its square was separable, and that a certain subspace of it was not separable. I had to stare at it for about a minute before I understood why I had not just proved a contradiction. – gowers – 2010-05-05T09:18:40.970

24Perhaps some discontinuities between metric space theory and topology arise because when studying metric spaces, distances are mysteriously required to be symmetric, and the requirement is dropped when switching to topological spaces. So you can have a point x that has y as a limit (i.e. it is in all neighbourhoods of y) but y doesn't have x as a limit. With this idea in mind, you can make any topological space separable by adding a single point, and making it belong to all open sets. – Marcos Cossarini – 2010-05-05T14:53:26.463

2I made this mistake once during a seminar talk. I thought I could just intersect the dense subset with the subspace. Luckily, I could correct it since everything was metrizable :D – Martin Brandenburg – 2010-05-05T22:50:05.743

3In a similar spirit: every subgroup of a finitely generated group is finitely generated. – Terry Tao – 2010-06-06T18:19:53.057

Terry, that belongs to a whole new line of false beliefs: "A submodule of a $k$-generated module is $k$-generated, e.g. a submodule of a cyclic module is cyclic." Corollary: a left ideal in any ring is principal. Or "A subring of a finitely generated ring is finitely generated", etc. – Victor Protsak – 2010-06-10T08:19:17.140


Here are a few more: (Everything between quotation marks is a false belief.)

Basic logic: Among students: "If A implies B then B implies A" (or "if A implies B then not A implies not B").

Even among mature mathematicians a frequent false belief is to forget that the conclusion of a theorem need not hold once the conditions of the theorem fail. Another common frequent belief is to assume that once the conditions fail then the conclusion must fail too.

Calculus: "The derivative of a differentiable function is continuous."

"An infinite series whose general term tend to 0 is convergent."

Geometry: "The circle is the only figure which has the same width in all directions." (Feynman regarded this mistake as one reason for the space shuttle Challenger disaster).

Polytopes: Often people believe that "given a convex polytope P you can slightly move the vertices to rational positions keeping the structure of the polytope unchanged."

(From Udi de Shalit): Some people believe that "if you hold a cube along a main diagonal, the remaining vertices all lie on a plane." Some even say that their number is 4.

Algebra (Also from Udi) "I have encountered many misconceptions about solvability by radicals. Some people think that 'the solution of an irreducible equation of degree 5 and higher, say over $\mathbb Q$, is never expressible by radicals'. Some amateur mathematicians even say that 'equations of degree 5 and higher have no solutions'."

Probability: "If you play the casino patiently and carefully you will win in the long run" (and "you do not believe that?, this is my own experience on the matter!" and "Indeed when I am calm and patient I win, but when I lose my temper I lose big time".)

"an event which may occur has positive probability": (not true for infinite probability spaces)

Various places: "If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures."

Combinatorics: "This is a finite problem, surely you can solve it with a computer."

"Hall marriage theorem is very nice and I am surprised no combinatorialist bothered to extend it to a matching built from triples instead of pairs." (It is unlikely that a general characterization when a hypergraph built from triples has a perfect matching (of triangles) will be found.)

Computer science: "It is known that quantum computers can solve NP complete problems in polynomial time."

Gil Kalai

Posted 2010-05-04T21:02:58.510

Reputation: 12 545

2i love the example, but i must challenge Feynman's discovery – Mio – 2011-03-04T02:50:30.093

1@JoelDavidHamkins: the statement which falsehood was asserted wasn't "quantum computers can solve NP complete problems in polynomial time"; it was "It is known that quantum computers can solve NP complete problems in polynomial time." :) – Michael – 2013-09-18T21:14:05.513

About Computer science: "It is known that quantum computers can solve NP complete problems in polynomial time." A wrong believe is rather that quantum computers can calculate more things than classic computers; which is false. quantum calculability is equivalent to classic calculability. – Philippe Gaucher – 2013-12-03T08:30:40.507

As a variation on the last one, I keep hearing that it is known that quantum computers can quickly solve problems that require exponential time on classical computers. A related common belief is that NP-complete problems have exponential lower bounds. – Theodore Norvell – 2014-06-04T14:12:10.497

http://www.cut-the-knot.org/do_you_know/cwidth.shtml – kjetil b halvorsen – 2015-03-05T15:16:55.150

15"If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures." This is not necessarily false. In some important cases this does work! One uses the compactness theorem for such proofs. – Johannes Hahn – 2010-05-07T12:46:25.127

The last one surely is related to the (wrong) expansion for NP as non-polynomial. I am, however, optimistic that the reason for that particular error has to do with BQP and QMA classes being less publicized in popular literature. Wait a few more years and this mistake may die out. – Willie Wong – 2010-05-07T14:00:05.653

1Johannes, many false beliefs can be rescued in some important scenarios; Willie, what I simply wanted to say is that some people believe that every NP problem can be solved in polynomial time by a quantum computer and this is not known to be the case, and, in fact, there is evidence to the contrary. – Gil Kalai – 2010-05-07T15:00:48.993

11Probability: There are two opposite errors. Both are common. Say we are flipping a fair coin repeatedly. (1) if there have been more heads than tails, then tails is "overdue" and thus more likely on the next flip. (2) if there have been more heads than tails, then heads is "hot" and thus more likely on the next flip. – Gerald Edgar – 2010-05-07T15:15:10.267

1The better question about Hall is whether there is a version for tripartite graphs (which is merely a weak particular case of the hypergraph version). Are there results on it? – darij grinberg – 2010-05-09T12:27:03.920

How about this? http://mathoverflow.net/questions/22901/tripartite-graph

– JBL – 2010-05-09T15:11:21.903

17I don't know anything about polytopes, but I'm having a hard time disbelieving this false result. Are we talking about finite polytopes here? – Tom Ellis – 2010-05-09T19:26:44.237


Entirely finite, Tom. There are 4-dimensional polytopes with 33 vertices that cannot be presented with rational coordinates. Here is a reference http://arxiv.org/PS_cache/arxiv/pdf/0710/0710.4453v2.pdf

– Gil Kalai – 2010-05-09T20:46:06.260

3@GeraldEdgar, (responding to a 5 year old comment but...) Then there is the false belief that if you toss a coin 1000 times and you get 1000 heads, then the next toss will be 50/50 heads/tails (I never said that the coin was fair!). These are the Gambler's Fallacy and the Gambler's Fallacy Fallacy respectively. – fhyve – 2015-04-18T05:53:35.343

1Well that is very interesting, and counterintuitive! – Tom Ellis – 2010-05-10T17:35:04.630

34You list the statement "Quantum computers can solve NP complete problems in polynomial time" as a false belief, but I don't believe you actually know this belief to be false. For example, the assertion that this belief is false implies $P\neq NP$. Perhaps the false belief that you intend to mention is: "It has been proved that Quantum computers can solve NP complete problems in polynomial time." – Joel David Hamkins – 2010-05-17T12:34:16.570

Right this is more or less what I meant. – Gil Kalai – 2010-05-17T12:49:04.673

1"when I am calm and patient I win, but when I lose my temper I lose big time" $-$ wasn't that the purported "system" of Dostoyevsky's Gambler? – Noam D. Elkies – 2015-09-05T23:36:32.577

Interesting! I read the book long time ago but I don't remember. I think I heard from people who gamble the claim that when they gamble carefully and not too recklessly they win in the "long run". – Gil Kalai – 2015-09-08T16:34:16.553

2Happy to give this answer its 100th vote. This rational polytope black magic is the most unbelievable thing I've learned since $5 \times 12 = 60$ (a long time ago). – Maxime Bourrigan – 2016-01-29T17:17:30.220

There are some interesting mathematical issues related to the casino example, such as the false belief that $\mathbb{E}(S_\tau)=0$ if $S_n=X_1+X_2+\ldots+X_n$ is the sum of independent random variables with zero expectation and $\tau$ is an (a.s. finite) stopping time. This "disproves" that the strategy of leaving the table when you've won enough may work, but falls victim of the martingale bet. Ironically, $S_n$ is the best known example of a martingale. – Victor Protsak – 2010-06-10T07:33:58.973

1@Gerald: I think the first false belief is much more common among learned scholars: if one had sat in 1000 airplanes and not experienced a crash yet, then the next one will very likely be a crash. – John Jiang – 2010-10-10T19:42:57.710


I used to believe that a continuous algebra homomorphism from $k[[x_1,\dots, x_m]]$ to $k[[y_1,\dots,y_n]]$, with $m > n$, could not be injective. Konstantin Ardakov set me straight on this.


Posted 2010-05-04T21:02:58.510

Reputation: 14 949

@fherzig $f$ is the root of the monic polynomial $t-f(v)\in k(v)[t]$. What are you actually want to say? – user26857 – 2015-07-08T16:48:42.960

32Whoa, that's not true?! What's the counter-example? – David E Speyer – 2010-05-21T19:44:02.273

@user26857 That doesn't work, because $f(v)$ is an element of $k[[v]]$, and not necessarily $k(v)$, – Wojowu – 2015-10-17T19:46:01.863

26Not true and deeply frustrating. Take a map from $k[[x,y,z]]$ to $k[[u,v]]$ that sends $x$ to $u$, $y$ to $uv$ and $z$ to $uf(v)$ for some $f\in k[[v]]$ and think about what the kernel is. It isn't hard to see that only for countably many choices of $f$ can it possibly be zero. – Simon Wadsley – 2010-06-07T19:45:47.347

4@Simon: Were you thinking of $k$ a countable field, and did you mean "can it possibly be non-zero"? (It seems the condition on $f$ is that it should not be a solution of any monic poly. whose coefficients are in $k(v)$.) – fherzig – 2010-06-07T22:00:17.407

3Yes. I had just come back to say that I was assuming that $k$ is countable (in fact in my mind I was thinking of $k$ as finite because that is the case that I think about most) and as you say I should have said 'can it possibly be non-zero'.

Of course the countability of $k$ is not necessary for such an example it just makes it easy to see that a suitable $f$ exists. – Simon Wadsley – 2010-06-08T08:25:32.560

10Thanks, Simon, I had meant to get around to coming back and answering David! It's disturbing, isn't it? I had a paper ready to go apart from one lemma I was confused about that used needed this (false) statement, and indeed, not only is the statement wrong, the thing I thought I had almost proved turned out to be false! – JSE – 2010-06-08T22:15:33.327

Indeed, thank you Simon! – David E Speyer – 2010-07-06T16:08:39.020

Reading all of these has made me really glad that I am... "slower" than most people. To make all of these errors you would at least need to have a little intuition or understanding as to why they should be plausible. When I started learning algebraic geometry I was so lost and understood so little that I rigorously proved every little detail of every claim I made because I couldn't even figure out which ones were obvious enough not to prove completely and didn't want to look like a ding-dong in front of my advisor. – Prince M – 2017-11-19T05:17:16.553


$$2^{\aleph_0} = \aleph_1$$

This is a pet peeve of mine, I'm always surprised at the number of people who think that $\aleph_1$ is defined as $2^{\aleph_0}$ or $|\mathbb{R}|$.

Amit Kumar Gupta

Posted 2010-05-04T21:02:58.510

Reputation: 2 544

41Show me a proof that it's not! – I. J. Kennedy – 2011-06-15T21:39:45.383

10Or: Show me a proof that it is correct! ;-) – Martin Brandenburg – 2011-10-03T07:25:20.517

15This error can be found in Gamow's famous book "One, Two, Three...Infinity", and even the Oxford English Dictionary contains a quote in their definition of "aleph null":

"There is no infinite number between aleph-null (the number of positive integers) and aleph-one (the number of real numbers)."

(apparently pulled from a Scientific American article) – Todd Eisworth – 2013-04-16T18:38:05.910

1Please tell me why this axiom is a false belief? – Anixx – 2015-01-09T12:06:11.410

14The key word is "defined". The false belief that annoys me is that people believe $\aleph_1$ is defined as $|\mathbb{R}|$. – Amit Kumar Gupta – 2015-01-09T16:14:47.260

9(This also appears in "The gods themselves" by Asimov but, it being science-fiction, I always "fix it" by imagining that in the world of the story, $\mathsf{CH}$ has been accepted as an axiom.) – Andrés E. Caicedo – 2015-01-13T22:01:18.553


I think, there are different types of false beliefs. The first kind are statements which are quite natural to believe, but a moment of thought shows the contradiction. Of this type is the sin-example in the opening post or a favorite of mine (also occured to me):

  • The underlying additive group of the field with $p^n$ elements is $\mathbb{Z}/p^n\mathbb{Z}$.

The other type is also quite natural to believe, but one has really to think to construct a counter example:

  • Every contractible manifold is homeomorphic to $\mathbb{R}^n$.
  • Every manifold is homotopy equivalent to a compact one.
  • Quotients commute with products in topological spaces.
  • Every connected component of a topological space is open and closed. Or related to this:
  • To give a continuous action of a topological group $G$ on a discrete space $X$ is the same as to give an action of the group of connected components of $G$ on $X$.

Lennart Meier

Posted 2010-05-04T21:02:58.510

Reputation: 5 879

Is it really that hard to construct a topological space with non-clopen connected components, or has looking at profinite groups too much jaded me? – Jonathan Kiehlmann – 2010-12-01T19:43:30.927

2Profinite groups are not the topological space, most topologists are most familiar with. – Lennart Meier – 2010-12-03T14:42:33.570

1What contractible manifolds are there besides $\mathbb{R}^n$? I can only think of if you add boundary and get the closed unit ball or a half space or something a little more complicated... Are there completely different examples? – Olivier Bégassat – 2011-03-25T09:10:38.210

2http://en.wikipedia.org/wiki/Whitehead_manifold – Lennart Meier – 2011-03-25T23:24:12.993

3@Lennart: True, but topologists know the rationals. They are (usually) not the ones with this mistaken belief. – Gerald Edgar – 2011-05-04T13:14:54.717

The finite fields bit hits very close to home, especially when you start denoting them $\Bbb F_q$, because now the $p$ is gone from your notation, so it's very easy to forget that additive group isn't just cyclic of order $q$. – Dylan Yott – 2014-02-11T15:13:43.863

37Along the same vein as the first example: the field with $p^2$ elements is a subfield of the field with $p^3$ elements (etc...) – Sean Kelly – 2010-05-05T17:06:18.750

1"Every manifold is homotopy equivalent to a compact one" has an easy counterexample. Infinitely many copies of your favourite manifold. – Manuel Bärenz – 2015-03-03T14:55:03.143

Though if I remember correctly, the field with 2^2 elements is a subfield of that with 2^4 elements, which is a subfield of that with 2^8 elements, and so on. – Tanner Swett – 2010-05-11T00:23:13.337


From the Markov property of the random walk $(X_n)$ we have

$$P(X_4>0 \ |\ X_3>0, X_2>0) = P(X_4>0\ |\ X_3>0).$$

To paraphrase Kai Lai Chung in his book "Green, Brown, and Probability",

"The Markov property means that the past has no after-effect on the future when the present is known; but beware, big mistakes have been made through misunderstanding the exact meaning of the words "when the present is known"."


Posted 2010-05-04T21:02:58.510


It may easily happen that $X_2 X_3 <0$ always, so LHS probability may be even not defined. – Fedor Petrov – 2016-04-20T19:18:10.917

@JohnJiang If ${wk}{k}$ is a (discrete) Markov process on a filtered probability space $(\Omega, \mathcal{F},{\mathcal{F}k}{k}, \mathrm{P})$ and we define $x_{k+1} = x_k + w_k$, then we may say that ${xk}{k}$ is driven by ${w_k}k$, but it is not a Markov process itself. Indeed, $$\mathrm{P}[x{k}=\alpha\mid x{k-1}=\beta, x{k-2}=\gamma] = \mathrm{P}[w{k-1}=\alpha-\beta\mid w{k-2}=\beta-\gamma]$$ which is different from $\mathrm{P}[x{k}=\alpha\mid x{k-1}=\beta]$. But, $(x_k, w_k)$ is a Markov process. – Pantelis Sopasakis – 2016-06-25T15:41:23.500

14This is a nice one. I almost fell for it. The best way to see it's not true is perhaps to condition on $X_3> 0, X_2 \le 0$. Then that forces X_4 < 1, if the random walk has increment $\pm 1$. – John Jiang – 2010-10-10T19:40:06.303


Here are two group theory errors I've seen professionals make in public.

1) Believing that if $G_1 \subset G_2 \subset \cdots$ is an ascending union of groups such that $G_i$ is free, then $\bigcup_{i=1}^{\infty} G_i$ is free. Probably the vague idea they have is that any relation has to live in some $G_i$, so there are no nontrivial relations.

2) Consider a group $G$ acting on a vector space $V$ (over $\mathbb{C}$, say). Assume that $G$ acts as the identity on a subspace $W$ and that the induced action of $G$ on $V/W$ is trivial. Then I've seen people conclude that the action of $G$ on $V$ is trivial. Of course, this is true if $G$ is finite since then all short exact sequences of $G$-modules split, but it is trivial to construct counterexamples for infinite $G$.

Andy Putman

Posted 2010-05-04T21:02:58.510

Reputation: 28 809

1for 2), it is also easy to build counterexamples for finite $G$ with order divisible by $p$ and a vector space over the field $\mathbb{F}_p$: take for instance $G = \mathbb{Z}/p\mathbb{Z}$ and let it act via the matrix $\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$ over $\mathbb{F}_p$. It is true if $G$ is finite only if the vector space is over a field of characteristic prime with $|G|$. – Maurizio Monge – 2011-03-06T14:13:29.070

@Maurizio Monge : Notice that I wrote ``over $\mathbb{C}$''. – Andy Putman – 2012-02-12T20:18:39.057

5Wait, what ascending chain of free groups is $\mathbb{Q}$ the union of? – ziggurism – 2012-11-28T05:27:31.043

@JoeHannon: Use $G_i = { n2^{-i} | n \in \mathbb{Z} }$. – yatima2975 – 2013-03-11T20:45:14.057

11@yatima You probably mean $G_i={n/i!}$. – The User – 2013-05-11T23:58:43.970

31So people think $\mathbb{Q}$ is a free group? Curious. – Pete L. Clark – 2010-05-05T23:39:02.223

29Sadly enough, I suspect that many people who care about geometric/combinatorial group theory do not think of Q as a group... – Andy Putman – 2010-05-06T00:28:42.357

14Pete, they don't think... they BELIEVE! – Victor Protsak – 2010-05-14T06:56:42.180

6A nice group-theoretic one I've seen often is that if $A$ is any abelian group, then the torsion subgroup is a direct summand. – Steve D – 2010-05-15T13:39:51.727

16Come on Steve -- aren't all abelian groups finitely generated? =) – Andy Putman – 2010-05-15T15:04:49.840


The field of $p$-adic numbers has characteristic $p$.

Yiftach Barnea

Posted 2010-05-04T21:02:58.510

Reputation: 3 525

4This seems to be an artifact of notation, since things with subscript $p$ have characteristic $p$ in early algebra. I've seen people make this mistake while simultaneously holding the (correct) beliefs that a) $\mathbb{Q}\subset \mathbb{Q}_p$ b) $\mathbb{Z}$ is dense in $\mathbb{Z}_p$ and c) $\mathbb{Z}_p/p\mathbb{Z}_p \cong \mathbb{Z}/p\mathbb{Z}$

Any one of which would be evidence against the false belief. – David White – 2011-05-03T19:44:49.803

1Related: The unit sphere in the $p$-adics has measure zero. – Phil Isett – 2011-07-03T19:17:31.010

This one I heard from a reputed colleague in another field, but I presume he didn't really think of it and only got influenced by the notation. – ACL – 2016-04-21T14:41:57.047


I remember from my first analysis class thinking that if $\mathbb{Q}\subset E\subset\mathbb{R}$ with $E$ open, then $E$ would have to be all of $\mathbb{R}$ (at least more or less, maybe up to countably many points). And once we started measure theory I remember arguing with a friend over it for a good two hours.

Owen Sizemore

Posted 2010-05-04T21:02:58.510

Reputation: 1 543

2This was one of the most startling realizations I have had mathematically. I was really shocked when I first thought about this (also in a measure theory class). – Steven Gubkin – 2010-05-05T18:39:30.500

50The construction of a counterexample is really not too difficult: enumerate Q and take the union of intervals of size 1/3^n about the nth rational. Morals: Q is small and open sets are weird. – Qiaochu Yuan – 2010-05-06T00:49:15.903

2That is the best kind of counterexample: one that is just completely obvious after someone points out that there is any problem. – Steven Gubkin – 2010-05-06T01:07:05.067

3This is indeed the best kind of counterexample, but gowers has actually mentioned it in the original question. – Zsbán Ambrus – 2010-05-06T06:49:43.180

10@Qiaochu Yuan: That example as an indicator function is also what I use to really illustrate the power of Lebech integration over Riemann integration. The usual example of 0 if x is irrational and 1 if it is rational leaves me unsatisfied since later one proves that lebech integration only depends on the equivalence class of the function and under this equivalence this function is 0. – Thomas Kragh – 2010-05-06T08:58:16.093

12Where do you get the spelling "Lebech"? I've only ever seen "Lebesgue," and the only "Lebech" in Wikipedia is a contemporary Danish politician. – Nate Eldredge – 2010-05-06T14:43:37.143

@Nate: I venture to guess that it's a phonetic-based slip of mind. – Victor Protsak – 2010-05-06T16:12:55.680

1Consider $E=\Bbb R\setminus{x:x\text{ is expressible as $0.a1b10c100d1000\dots$ in binary}}$. This is open (since its complement contains its limit points), contains the rationals, has uncountable complement, and doesn't need any measure theory to understand. – Akiva Weinberger – 2015-08-31T17:07:45.837

37Measure theory? Enumerating rationals? Isn't $\mathbb{R}-{\sqrt{2}}$ open and contains the rationals?? – Dror Speiser – 2010-05-20T15:30:11.130

28Dror, note the parenthetical remark in Owen's post. – Kevin H. Lin – 2010-05-25T21:21:53.717


"Either you can prove the statement, or you can find a counterexample."

This statement is usually applied to universal statements, those having the form $\forall x\ \varphi(x)$, where the concept of counterexample makes sense, but the general sentiment is the belief that every statement in mathematics is either provable or refutable.

The belief is false, because of the independence phenomenon.

Joel David Hamkins

Posted 2010-05-04T21:02:58.510

Reputation: 152 193

8To counter this false belief, exam questions should always start with:

Prove, disprove or prove that neither is possible. :) – user11235 – 2011-04-10T11:37:08.797

35Thei, this would merely put off the dilemma one more step, since perhaps the statement is neither provable, nor refutable, not is its resulting independence provable! So one would need to add a fourth possibility, "or prove that none of the preceding is possible", and then repeat this process infinitely many times, with smaller and smaller fonts on the exam sheet. – Joel David Hamkins – 2011-10-06T20:50:38.083

There are some theorems of the form "either something is true and there is a proof, or it is false and there is a counterexample". In my opinion, when it does not hold, it is because those notions of "proof" and "counterexample" are not good enough for that purpose – user49822 – 2014-07-25T18:35:01.037

10Goedel's completeness theorem says that under certain hypotheses, there's either a proof or a counterexample. But "under certain hypotheses" needs to be understood. – Michael Hardy – 2010-06-06T19:50:31.550

3For people whoc categorize people in two classes: there are two kinds of mathematicians. Those, optimistic, who prove statements. Those, pessimistic, who construct counter-example. Which one are you ? – Denis Serre – 2010-09-30T15:42:58.777


If $f(x,y)$ is a polynomial with real coefficients, then the image of $f$ is a closed subset of $\mathbb{R}$. Note. Problem A1 on the 1969 Putnam exam asked to describe all possible images of $f$. I was told that the writers of this problem did not realize its subtlety.

Richard Stanley

Posted 2010-05-04T21:02:58.510

Reputation: 26 417

68One concrete counterexample: $(1-xy)^2+x^2$ – Richard Dore – 2010-07-08T05:35:49.857


In order to show that a polynomial $P \in F[x_1,\ldots,x_n]$ vanishes, it suffices to show that $P(x_1,\ldots,x_n) = 0$ for all $x_1,\ldots,x_n \in F$. True in infinite fields, but very false for small finite fields.

Closely related: if two polynomials $P$, $Q$ agree at all points, then their coefficients agree. Again, true in infinite fields, but false for finite fields.

(This is ultimately caused by a conflation of the concept of a polynomial as a formal algebraic expression, and the concept of a polynomial as a function. Once one learns enough algebraic geometry to be comfortable with concepts such as "the $F$-points $V(F)$ of a variety $V$" then this confusion goes away, though.)

Terry Tao

Posted 2010-05-04T21:02:58.510

Reputation: 53 985

This seems related to the widely held belief that 1+1=2, (and how could 2 be equal to 0 ?!). On a psychological level this may connect with the assumption we all tend to make that a is different from b when asked to count the elements of set {a,b} (a related entry is somewhere else on this page), or that in general two different symbols are instinctively taken to denote two distinct objects. – Vania Mascioni – 2010-06-06T20:53:08.580

This is another example of intension/extension (the distinction I was trying to get at in question http://mathoverflow.net/questions/18848/extensional-theorems-mostly-used-intensionally).

– Jacques Carette – 2010-06-08T00:04:32.333

Also closely related: Fermat's Little Theorem can be used to prove equality of polynomials. E.g., $X^2=X$ in $\mathbb{Z}_2[X]$ because $x^2=x$ in $\mathbb{Z}_2$. – Jose Brox – 2016-04-22T08:20:10.920


"Automorphisms of the symmetric group $S_n$ are inner (that is, each one is of the form $x \to axa^{-1}$ for some $a \in S_n$)" is a popular misconception, false for nontrivial reasons when $n=6$. That is both an easy mistake to make and important conceptually as an early hint of the complexities and special combinatorics that arise in finite group theory. Many people make it through a first class in group theory without understanding that something different happens for $S_6$ and in doing so have missed an important piece of the the big picture, as far as finite groups are concerned.

It is easy to implicitly or explicitly acquire this belief, because:

  1. those really are all the automorphisms for $n$ other than 6, and

  2. the inner automorphisms are used so often, for all values of $n$ (or $n>2$) without distinguishing any specific case as unusual.

  3. $S_n$ behaves in many ways as a family of similar groups rather than a list of individual groups with their own diverse features. A typical proof might show some property of $S_n$ by induction on $n$, starting from a small value such as $n=1$ for basic properties, or $n=3$ to assure noncommutativity. Apart from the classification of symmetric group automorphisms itself (exposure to which would be an explicit articulation and correction of the false belief), these arguments never start as high as $n=7$ and I don't know of any that distinguish $n=6$ or some equivalent case as a lone nontrivial exception. So it is easy to get the idea of more uniformity in the $S_n$ than really exists.

In essence, there are no obvious clues in the environment that $n=6$ might be special, and a number of indicators that no special case should exist at all.


Posted 2010-05-04T21:02:58.510

Reputation: 3 042

@NateEldredge : "Combinatorial Structure of the automorphism group of $S_6$" by T.Y. Lam and David B. Leep, Expositiones Mathematicae 11 (1993), no. 4, pp. 289--903. $$.$$ In this paper, it is claimed that the group of all $1440=6!\times 2$ automorphisms of $S_6$ has exactly three distinct subgroups of index 2, no two of which are isomorphic to each other. (One of those is of course the group of all $6!=720$ inner automorphisms.) $\qquad$ – Michael Hardy – 2016-03-10T04:55:52.973

4Thanks for that example -- and the detailed diagnosis that accompanied it. – gowers – 2010-06-10T20:23:22.247

3The main reason why one can fall for this is that any two permutations in $S_n$ with equal cycle type (and thus, any two permutations in $S_n$ which can be automorphed into each other) are conjugate, so one would expect that the same holds "globally". – darij grinberg – 2010-06-10T21:53:56.607

6Cycles presume a given representation (group action) of $S_n$. The defining representation is permutations of an $n$-point set; "$S_n$-automorphisms arise from permuting the points" is a restatement of the false belief, and so cannot fully explain it.

$S_n$ as abstract group carries only its regular (Cayley) representation, permuting a set of size $n!$. The possibility of non-conjugate elements in $S_n$ having the same cycle type in the regular representation opens the door for a nontrivial outer automorphism to exist. Missing these ideas may be the origin of the error, in "cycle" terms. – T.. – 2010-06-10T23:31:41.077

To clarify: this is true for all n other than n=6? Can you describe, or give a reference for, a non-inner automorphism of $S_6$? – Nate Eldredge – 2010-06-11T12:53:54.357

3@Nate: The outer automorphismen is induced from swapping the conjugation class of all transpositions (ab) with all tripeltranspositions (ab)(cd)(ef). – Johannes Hahn – 2010-06-11T14:30:19.560


Nate, automorphisms for $n=3,4,5$ and $n /geq 7$ fix the conjugacy class of transpositions (because no other class has the same size), thus are inner. The outer automorphism of $S_6$ is unique up to conjugation and several constructions are described at: http://en.wikipedia.org/wiki/Automorphisms_of_the_symmetric_and_alternating_groups

– T.. – 2010-06-11T18:06:35.240

( $n \geq 7$ ... ) – T.. – 2010-06-11T18:07:22.370


"If any two of the $3$ random variables $X,Y,Z$ are independent, all three are mutually independent." In fact, they may be dependent; the simplest example is probably $(X, Y, Z)$ chosen uniformly from $\{(0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1)\}$.


Posted 2010-05-04T21:02:58.510

Reputation: 1 378

This is a false belief? – Harald Hanche-Olsen – 2010-05-05T00:26:17.003

Harald, I've edited to clarify. – JBL – 2010-05-05T00:42:06.283

46or in words: take two fair $0-1$ coins and the random variables ""value first coin", "value second coin" and "sum modulo 2". – Alekk – 2010-05-05T08:25:04.640

I like the variant: two fair $\pm 1$ together with their product. – Mark Meckes – 2010-05-05T12:59:35.023

13A physics professor assigned proving this false statement as homework in a statistical mechanics course. – Douglas Zare – 2010-05-05T17:43:34.497


Some people have trouble understanding that (and why is) 0.999... = 1


Posted 2010-05-04T21:02:58.510

Reputation: 8

3Many mathematicians will explain $.999\ldots=1$ in terms of infinite series and limits, thus taking the reals as Cauchy completion of the ration as the implicit background. No surprise that they win few converts.

Much better pedagogically to view $x=.999\ldots$ as the infinite conjunction $x\in[0,1]$ and $x\in[.9,1]$ and $\ldots$ and observe, on geometrical grounds, that the intersection, which obviously contains 1, must consist of a single point (thus taking the Dekekind completion of the rationals as the implicit background). – David Feldman – 2010-12-17T23:35:42.760

The reals should serve as the place where one learns about limits - we should not the limit concept prerequisite to an honest conversation about the meaning of the numbers themselves. – David Feldman – 2010-12-17T23:35:47.550

3Or you could just tell them x = 0.999... 10 x = 9.999... 9 x = 9 x = 1 – person – 2011-01-17T11:07:44.867

26Another argument that I have sometimes found effective is to get people to agree that 0.33333... = 1/3 and then multiply both sides by 3. But sometimes this just causes people to believe that 0.3333... is not quite equal to 1/3. – Timothy Chow – 2011-03-24T14:18:31.893

2"many don't think that 0 is an even number"

Including a physics professor who taught an (otherwise very good) physics intro course and emphasized the parity problem with the phrase:

"The question of whether 0 is even or odd is like the question of whether hydrogen is a noble gas." – user11235 – 2011-04-10T11:42:18.310

3one of the worst gifts you can give a person is convincing them of this fact. they will never again be able to communicate with their peers and i have heard some may lose their jobs as math teachers. – roy smith – 2011-04-14T19:11:34.643

10A somewhat glib "constructivist" argument that might work on some people: Suppose I have a number 0.999...; you say it is not 1, so what would you have to add to my 0.999... to make it 1? When they say 0.000..., point out that if I never stop adding 9-digits to my number, then they can't stop adding 0-digits to theirs, so surely theirs is 0, and mine is 1. – Chris Pressey – 2012-12-01T13:07:25.477

146And as Ehud de Shalit mentioned to me, some once they understand that 0.9999... = 1 think that 0.8888888...= 0.9 – Gil Kalai – 2010-05-05T18:44:36.083


That's great Gil! And according to this copiously referenced article, http://en.wikipedia.org/wiki/Parity_of_zero, many don't think that 0 is an even number

– Victor Protsak – 2010-05-05T23:37:22.083

2http://en.wikipedia.org/wiki/0.999...#Infinitesimals is a great mathematical explanation for the troubles. – Martin Brandenburg – 2010-05-06T00:50:26.347

1But they might be indeed different! See, for example, [C. Schneider, R. Pemantle. When is 0.999... equal to 1?. Amer. Math. Monthly 114 (2007) 344-350.]. – Wadim Zudilin – 2010-05-07T12:33:59.830

34@Victor: Many don't think that 0 is even a number. :) – Wadim Zudilin – 2010-05-07T12:35:50.950

18Those that think 0 is a number sometimes don't think that -1 is a number, and of the latter, some don't think that i is a number. I'm not sure if I think that j and k are numbers. – Tom Ellis – 2010-05-09T19:16:01.947

7I have non-math friends who still don't believe me about this. – David Carchedi – 2010-05-11T17:30:53.100

I had trouble understanding if and why floor(0.999...) = 10. – Houshalter – 2016-05-02T08:51:35.117


There are a couple of false beliefs regarding the $I$-adic completion functor, where $I$ is an ideal in a commutative ring $A$.

The first is that the completion of an $A$-module $M$ is complete, or in other words, that the completion functor is idempotent. This is true if $A$ is noetherian, but false in general. I find this quite unexpected - you take a module, "complete" it, and the result is not complete...

Another issue is the exactness of the completion functor. The completion functor is exact on the category of finitely generated modules, but when you consider arbitrary modules, even over a noetherian ring, it is neither left-exact (this is easy to see) nor it is right-exact (this probably less known).

the L

Posted 2010-05-04T21:02:58.510

Reputation: 645

27Oh my god .... – Martin Brandenburg – 2010-05-19T14:18:19.633

Can you give an example of an $A$-module $M$ with non complete completion? Thank you! – Ricky – 2010-07-23T11:53:16.740


@Ricky, see for example the paper http://arxiv.org/abs/0902.4378 - example 1.8.

– the L – 2010-07-25T10:23:20.077


"It is impossible in principle to well-order the reals in a definable manner."

To be more precise, the belief I am talking about is the belief that well-orderings of the reals are provably chaotic in some sense and certainly not definable. For example, the belief would be that we can prove in ZFC that no well-ordering of the reals arises in the projective hierarchy (that is, definable in the real field, using a definition quantifying over reals and integers).

This belief is relatively common, but false, if the axioms of set theory are themselves consistent, since Goedel proved that in the constructible universe $L$, there is a definable well-ordering of the reals having complexity $\Delta^1_2$, which means it can be obtained from a Borel subset of $R^3$ by a few projections and complements. See this answer for a sketch of the definition of the well-order.

The idea nevertheless has a truth at its core, which is that although it is consistent that there is a definable well-ordering of the reals (or the universe), it is also consistent that there is no such definable well-ordering. Thus, there is no definable relation that we can prove is a well-ordering of the reals (although we also cannot prove that none is).

Joel David Hamkins

Posted 2010-05-04T21:02:58.510

Reputation: 152 193

7The best part of this answer is how the constructible universe subverts all the intuition you learn about AC from doing non-model theory. Experience would lead one to think that "AC = nonconstructive" in "the usual model of the real numbers", not realizing that there is no usual model. Your (my, everyone's) mental image of the reals is a sort of "lazy evaluation" (to use a programming term) of the model we would really like but haven't even specified fully. As you show in your answer, once given the facts we wouldn't even know which model that would be. – Ryan Reich – 2010-10-20T11:21:47.433

Your 'always' kept me wondering for a while! – Mariano Suárez-Álvarez – 2010-05-04T23:35:39.333

Yes, that is the nature of these independence results, isn't it? There is a definition that you can write down and prove that it is consistent that it defines a well-ordering of the reals, but it is also consistent that it doesn't. So does it or doesn't it? It depends on the set-theoretic background... – Joel David Hamkins – 2010-05-05T00:21:24.423

25I think that you are taking an imprecise statement and making it precise in such a way that it is wrong. – Marcos Cossarini – 2010-05-05T01:00:11.173

I think that there's no common confusion here. If you prove nonconstructively that such an order exists and then say that a constructive proof can't be given, everyone understands what you are saying. I wouldn't say that "it is possible to well-order the reals" only because a relation can be defined that can't be proved not to well-order of the reals. At the time in which undergrads learn about the existance of such a well-order, they are usually concerned on stating and then proving things, not in stating things that could be true in some model (they probably don't know about models). – Marcos Cossarini – 2010-05-05T01:06:21.640

7Marcos, the false belief I was aiming at was the belief that it is impossible in principle for a well ordering of the reals to be definable. That is indeed a false belief, since it IS possible in principle for there to be such a definable well-ordering, as there is one in Goedel's universe L. (Note: the constructible universe L has nothing at all to do with constructivism or constructive proofs, in the sense of your comment; Goedel used only classical logic.) And my remark was not aimed particularly at undergrads or even grads, but rather at research mathematicians holding that false view. – Joel David Hamkins – 2010-05-05T01:29:13.063

(In this case, "constructive" can be replaced by "not using AC") My point is that if someone says that "It is impossible to well-order the reals in a definable manner.", then it is most probable that he/she thinks that the well order cannot be proved to exist in ZF. But here the word "impossible" is tricky, you are using to quantify over the models, while I interpret as quantifying over the collection of existence proofs in standard set theory – Marcos Cossarini – 2010-05-05T03:31:32.757

So I would write the misbelief in a more precise way. – Marcos Cossarini – 2010-05-05T03:32:57.730

By the way, the constructible well order seems interesting. – Marcos Cossarini – 2010-05-05T03:41:21.890

1Marcos, I have edited. One paradox with the equation "constructive"="not using AC" is that in the constructible universe $L$, everything is constructible (in the sense of Goedel), but AC holds, and there are constructible well-orderings! – Joel David Hamkins – 2010-05-05T12:50:26.507


It's easy when you're an amateur to topology to assume any continuous bijection has a continuous inverse.The inverse of an arbitrary continuous bijection in a topological space is open,but it's NOT necessarily continuous. Continuity turns out to be a stronger condition.

The Mathemagician

Posted 2010-05-04T21:02:58.510

Reputation: 299

1This point is made very clearly by Mumford in section I.4 of his red book, (new version, p. 22; original p.42) by considering the (often false) assertion: "Bijective morphisms are isomorphisms". He presents several examples and counterexamples, and then in chapter III, discussing Zariski's main theorem, he presents the property of normality which rescues this intuition within algebraic geometry (ZMT). – roy smith – 2011-04-14T17:21:12.237

30Comments like this lead me to point the finger at some people's undergraduate education ;-) It was very common in my UG years to have "definition/basic properties/interesting counterexamples to plausible-sounding statements/example sheet with harder examples". For example, although it's not continuity but differentiability, I vividly remember being shown the map x-->x^3 on the reals very shortly after being told the definition of a diffeomorphism. – Kevin Buzzard – 2010-05-04T21:33:46.357

25Kevin's right. It's just as important to ask students for counterexamples as it is to tell them theorems. But I think this is hard for undergraduates in a way we tend to forget because we have slowly but surely rewired our brains to think in this way. When I was 20 I took a reading course from R. Narasimhan, and towards the end he told me to "Ask yourself the natural questions". Now, [harrumph] years later, this is second nature to me, but I remember at the time thinking that it was not so easy! – Pete L. Clark – 2010-05-04T22:28:46.437

6@Pete Coming up with good examples and counterexamples is one of the absolutely critical mathematical skills,not just for mathematicians,but good physicists and engineers. And you're right,it's something that really separates good programs from mediocre ones.Top programs have a lineup of top researchers and although sadly a lot of them leave a lot to be desired as teachers,the good teachers among them will insist on developing this skill in thier students.The best way to do that is to deemphasize full proofs and focus more on examples.A good example is worth more then a dozen full proofs. – The Mathemagician – 2010-05-05T05:51:11.207

@ Pete continued: I remember seeing the Cantor set for the first time in a measure and integration theory course with Gerald Itzkowitz and a lot of it's properties being given as one of the substantial problem sets in his notes. I went crazy with it,but boy,I learned a lot from those problems!!! – The Mathemagician – 2010-05-05T05:54:59.610

15I think one source of this problem are definitions (in the first lectures) like: a bijective morphism of groups is called an isomorphism. Introducing categories (very roughly!), defining the general notion of an isomorphism in a category and mentioning that it's awesome that for groups we just have to check bijectivity could really prevent this... – user717 – 2010-05-05T10:53:04.350

I'm lucky because our teacher taught us about continuous functions whose inverse isn't continuous in basic (univariate) calculus, before I even knew anything about topology. – Zsbán Ambrus – 2010-05-05T11:02:10.853

2A way to understand this is that the identity is not always continuous. – Gerardo Arizmendi – 2015-07-28T15:52:40.883

5I think the problem comes from the idea that once you solve something you're done. True, if it's your homework in undergrad technically I suppose you are, but SO MUCH can be learned from trying to prove the problem a different way or relaxing the conditions and seeing if your proof still holds (or could hold, or doesn't hold). Of course, being an undergraduate this is at least in part "Do as I say and not as I do" – Michael Hoffman – 2010-05-31T08:00:38.237

22There is at least two important cases where this intuition does hold, though: (a) when the continuous bijection is a linear transformation between Banach spaces (the open mapping theorem), and (b) when the bijection is from a compact space to a Hausdorff space. Of course, one only really appreciates these facts when one knows that the claim is false in general... – Terry Tao – 2010-06-05T21:13:36.203

4@Michael Hoffman: +1. Whoever came up with the expression "Beating a dead horse" was not a mathematician. On the contrary, taking a nontrivial result or example that you feel like you already know and turning it around and around again from different perspectives is a remarkably powerful technique for increasing one's mathematical knowledge. – Pete L. Clark – 2010-07-14T19:59:41.087

11...Some mathematicians have perfected this technique -- or possibly its converse? -- to the point where they become "venus fly traps": if an open problem flies too close to the domain of things that the mathematician understands ridiculously well, then...SNAP! That's the end of the problem. I have long admired Yuri Zarhin in this way, for instance. – Pete L. Clark – 2010-07-14T20:03:14.527


This is perhaps a misunderstood definition rather than a false belief, but:

"A subnet of a net $( x_\alpha )_{\alpha \in A}$ takes the form

$( x_\alpha )_{\alpha \in B}$ for some subset $B$ of $A$."

In truth, subnets are allowed to contain repetitions, and can be indexed by sets much larger than the original net. (In particular, there are subnets of sequences that are not subsequences.)

This false belief, incidentally, reinforces the false belief noted in a different answer, namely that compactness implies sequential compactness.

A precise Counterexample: The sequence $\sin(nx)$ is a sequence in the compact topological space $[-1, 1]^{\mathbb{R}}$ with product topology. So this net has a convergence subnet. But it is well known that the above sequence has no subsequence which is pointwise convergent (See the last page of the book of Walter Rudin's Principles of mathematical Analysis). So in this example the convergent subnet cannot be counted as a subsequence.

Terry Tao

Posted 2010-05-04T21:02:58.510

Reputation: 53 985

5Correcting that misunderstanding is crucial to prove that an accumulation point of a net is always the limit of some subnet, which does not hold for sequences. – Alfonso Gracia-Saz – 2010-10-19T06:26:31.120

I cannot find this (counter)example in Rudin's book (I have the 3rd edition). – Duchamp Gérard H. E. – 2016-12-25T06:32:46.810

It's Chapter 11, Exercise 16 on page 334 just before the bibliography in my copy of the 3rd edition of Rudin's Principles of Mathematical Analysis. – Dan Glasscock – 2017-01-03T22:57:14.193

I was cured of this one when I learned about compact spaces that have no convergent sequences other than eventually constant ones (why not take a sequence with distinct elements, pass to a convergent subnet, and there you have it ...?). – Christian Remling – 2017-08-22T01:47:34.283

Also the converse is false. The ordinal $\omega_1$ with the order topology (generated by the open intervals $(x,y)$ with $x\in\omega_1\cup{-\infty}$ and $y\in\omega_1\cup{+\infty}$) is obviously not compact, but it is sequentially compact: from any sequence $(xk)$ one can either extract a monotone decreasing or a monotone increasing subsequence $(x{k\ell})$, whose limit is $\bigcap x{k\ell}$ or $\bigcup x{k_\ell}$ respectively. Notice that in the second case the limit is a countable ordinal and thus does belong to $\omega_1$. – Mizar – 2017-09-16T14:57:17.987


Occasionally seen on this site: if a polynomial $P:\mathbb{Q}\rightarrow\mathbb{Q}$ is injective, so must be its extension to $\mathbb{R}$.

Yaakov Baruch

Posted 2010-05-04T21:02:58.510

Reputation: 1 816

23+1 Yaakov. $P(x) = x^3 - 5x$ is a counterexample. – Todd Trimble – 2011-03-31T21:47:35.537

1That's another funny one, because you would be very unlikely to see the same mistake with $\mathbb{C}$. So why are people liklely to make it with $\mathbb{R}$? – Thierry Zell – 2011-04-07T00:30:19.407

17@Tierry: continuity, of course. $\mathbb{R}$ is far from dense in $\mathbb{C}$ but $\mathbb{Q}$ is dense in $\mathbb{R}$. – Ryan Reich – 2011-04-22T18:06:08.370


Linear algebra: 1. If V is a vector space spanned by {ei} and W is a subspace of V then W is spanned by ek's contained in it. Actually, this is widely believed with bases in place of spanning sets. Or

2.   (U+V)∩W = U∩W + V∩W.

Both these "properties" are closely related to the current leader (by Tilman).

3. Every element of V⊗W is v⊗w with v∈ V, w∈ W.

All three are probably due to interpolating our intuition about sets to vector spaces.

4. Every symmetric matrix is diagonalizable.

Wait, didn't we prove this? ("True for the real matrices, so must be true in general").

Algebraic groups: if G is a linear algebraic group acting on a vector space V then the (Krull) dimension of the invariant ring satisfies the inequality

    dim k[V]G ≥ dim V-dim G,

or even a more precise belief that dim k[V]G=dim V-dim Gx for a generic x. This is true in the differentiable situation for the dimension of the quotient, when a compact Lie group acts smoothly on a manifold, and algebraic actions are "nicer", right?

Victor Protsak

Posted 2010-05-04T21:02:58.510

Reputation: 11 982

5>3 is a great example; I remember being caught off-guard by it indirectly even when I thought I was aware of it.< – Qiaochu Yuan – 2010-05-05T15:40:14.963

3(0 1;1 0) isn't diagonalisable over the field with 2 elements. It has 1 as an eigenvalue twice, but isn't the identity matrix. – Kevin Buzzard – 2010-05-05T20:22:02.563

@unknown: diagonalization by Gram-Schmidt involves dividing by the length (or think isotropic vectors) – Victor Protsak – 2010-05-05T21:02:51.873

2@Kevin: right, and if -1 is a square it's easy to construct a 3 by 3 nilpotent symmetric matrix – Victor Protsak – 2010-05-05T21:07:13.870

@unknown: That's where the usual proof by GS orthogonalization breaks down. "Isotropic" refers of course to the auxiliary positive-definite form used and you can construct a counterexample either that way or by making the matrix nilpotent (as David obligingly did for you elsewhere). – Victor Protsak – 2010-05-05T23:15:00.847

15>3 gets to play in physics-land, too --- the fact that there are non-simple tensors is the same as the fact that particles can become entangled in quantum mechanics.< – Matt Noonan – 2010-05-05T23:56:20.983

2@unknown. Just take ${\mathbb C}$. Every square matrix is similar to a symmetric one. Since there are non-diagonalizable complex matrices, some complex symmetric matrices are not diagonalizable. – Denis Serre – 2010-09-23T16:28:08.327


"The image of a category under a functor is a category."

This is a small one, but it lasted for six months when I was starting in category theory.

A finite counterexample exists, with just 3 objects. Even under a connectivity requirement, a small finite counterexample still exists. In fact, it is dead wrong to think anything like this holds.


Posted 2010-05-04T21:02:58.510

Reputation: 621

12You only need two objects 0, 1 and a single non-identity arrow 0->1. Consider the functor from this to Set which takes 0 and 1 to the natural numbers and the arrow to the successor function. – Todd Trimble – 2011-04-17T00:58:19.280

@Todd Trimble : Very nice! – Jérôme JEAN-CHARLES – 2011-04-20T10:34:02.023


People are silly. Did you ever notice how at airports, say, people happily walk around but when they come to a moving walkway, they tend to stop and take a break? Walking on a moving walkway is not any harder than walking on an ordinary walkway, and resting for 10 seconds here or there will get you to your destination 10 seconds later. So why do people stop for a rest in one place but hardly ever in the other?

The embarrassing bit is that I believed this logic myself for some time, and thought that people were indeed silly, until my son corrected me. I'm not sure if this falls under "a common false belief in mathematics", but it's certainly an amusing and confusing mistake to make.

Dror Bar-Natan

Posted 2010-05-04T21:02:58.510

Reputation: 1 400

15@ Sune: I think that Dror is suggesting, as the false belief, that people are silly. So he agrees with you. – Toby Bartels – 2011-04-04T07:48:06.310

3@Sune: Nice explanation.

Another possible way to think about it: Suppose there are several different moving walkways, some faster than others. Suppose further that you're required to rest for 10 seconds somewhere. On which moving walkway should you take your 10 second rest? Clearly on whichever one moves the fastest. – idmercer – 2011-07-09T22:57:46.850

This is so, so, so counterintuitive!! I love it. – Mehrdad – 2017-12-28T09:35:02.413


See also http://terrytao.wordpress.com/2008/12/09/an-airport-inspired-puzzle/

– Alison Miller – 2010-08-31T02:22:40.443

Amusing - thanks for the reference. --- Dror. – Dror Bar-Natan – 2010-08-31T10:35:56.530

37That is not a mistake: If you take a 10 sec break at the moving walkway you will arrive faster than if take a 10 sec break on the ordinary walkway. Imagine two people walking from A to B. One person take a 10 sec break just before the moving walkway the other take a 10 sec break as soon as he is on the walkway. They will begin to walk again at (almost) the same time, and the second person will have moved the distance the moving walkway moves in 10 sec. – Sune Jakobsen – 2010-09-01T09:13:05.733


That the notion of "picking a random number" is well-defined without providing any further information.

Ken Fan

Posted 2010-05-04T21:02:58.510

Reputation: 695

6In fact Knuth makes a similar joke in TAOCP chapter 3.1 (though he proves a different point with it). He asks whether 2 is a random number. – Zsbán Ambrus – 2010-12-04T22:54:03.870

3I've long suspected that the Dilbert comic alludes to the "Feynman point" where six 9's occur surpisingly early (before digit 1000) in the decimal expansion of $\pi$. – Noam D. Elkies – 2011-06-05T15:57:45.493

I find it intriguing that, if one interprets "random" as "Martin-Lof random", then in fact it is. – usul – 2015-10-17T19:09:44.260


Obligatory link: http://xkcd.com/221/

– Charles Stewart – 2010-07-13T10:20:09.607


And likewise http://dilbert.com/strips/comic/2001-10-25/

– Nate Eldredge – 2010-07-14T19:59:07.280


False belief: "There are no known sub-exponential time algorithms for NP-complete problems."

This one is tricky for a couple of reasons. The first is that the term "sub-exponential" is sometimes defined in different ways. With a sufficiently strong definition of "sub-exponential" the above statement is true, in the sense that there is no known separation of the complexity classes NP and EXPTIME (EXPTIME being the class of languages decidable in time $2^{p(n)}$ where $p(n)$ is a polynomial). However, it is quite common to refer to an $O(2^{\sqrt{n}})$ algorithm as "sub-exponential." It is trivial to construct an NP-complete problem that can be solved in sub-exponential time in this sense, because the standard definition of a reduction allows you to expand the size of the input from $n$ to $n^2$ say (e.g., by padding with zeros). But less artificial examples also exist, such as the planar traveling salesman problem, which was shown by Smith to be solvable in $2^{O(\sqrt{n})}$ time without any artificial padding. What is true is that there are many NP-complete problems, such as 3SAT, for which no subexponential algorithms are known if you do not artificially pad the representation of the instances. (Reducing 3SAT to planar TSP does not work because the instance size blows up during the reduction.)

Often this false belief shows up in the following form: "Factoring cannot be NP-complete because there are subexponential algorithms for factoring." It is true that factoring is not known to be NP-complete but the reasoning is wrong. Showing that factoring is NP-complete would not automatically yield subexponential algorithms for all other NP-complete problems.

Timothy Chow

Posted 2010-05-04T21:02:58.510

Reputation: 30 733

What a gem! I always thought this was true. – Mehrdad – 2017-12-28T09:36:41.570

9For clarity, the definition of sub-exponential for which this is true is $O(\mathrm{exp}(n^\epsilon))$ for all $\epsilon$. – Peter Shor – 2010-08-22T19:02:05.097


Some people believe there is no "formula" for the nth prime number. Of course there are many such formulas, even though not very useful: http://mathworld.wolfram.com/PrimeFormulas.html

The reason given for disbelieving the existence of a "prime number formula" is also curious: "because the primes are unpredictable". This belief is in contradiction with the simple fact that anyone can come up with an easy algorithm which gives the nth prime number. There is something mystical associated with this ill-defined term "formula".

Manuel Silva

Posted 2010-05-04T21:02:58.510

Reputation: 321


Doron Zeilberger's article in the PCM has an illuminating discussion in the very beginning about what constitutes an "answer" to a question like "what is the nth prime," albeit in the context of combinatorics: http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/enuPCM.pdf

– Qiaochu Yuan – 2010-06-08T14:10:18.567

See my comment to Vania Mascioni's answer, which I made before I saw your post. – Victor Protsak – 2010-06-10T07:04:50.043


It is a difficult open problem whether every finite group is isomorphic to the group of automorphisms of a finite extension $K$ of $\mathbb{Q}$. In fact, this result is true! See for instance http://www.jstor.org/pss/2043724. The actual "inverse Galois problem" also requires $K$ to be a Galois extension of $\mathbb{Q}$. The true theorem is equivalent to the statement that every finite group is isomorphic to a quotient group of a Galois group over $\mathbb{Q}$. I once observed a famous expert on algebraic number theory being confused on this issue.

Richard Stanley

Posted 2010-05-04T21:02:58.510

Reputation: 26 417

1Very interesting, I didn't know that result. – Martin Brandenburg – 2011-04-12T08:42:45.657


False belief: Saying that ZFC is consistent is the same as saying that if ZFC proves "there are infinitely many twin primes" (for example) then there really are infinitely many twin primes.

Everybody realizes that if ZFC is inconsistent then a formal ZFC proof of "there are infinitely many twin primes" tells us nothing about whether there really are infinitely many twin primes. A lot of people, without necessarily realizing it, turn this around and assume that the consistency of ZFC is the only condition needed to ensure that its theorems are "trustworthy." But this is not the case, even if we restrict our attention to first-order statements about the natural numbers. We say that ZFC is arithmetically sound if all first-order sentences about the natural numbers that are provable in ZFC are true. The arithmetical soundness of ZFC is a stronger condition than the consistency of ZFC. For example, Goedel's 2nd incompleteness theorem says that if ZFC is consistent, then ZFC doesn't prove "ZFC is consistent." So it's conceivable that ZFC is consistent but that "ZFC is inconsistent" is a theorem of ZFC. Then we would have an example of a theorem of ZFC that asserts something false, even though ZFC is consistent.

Timothy Chow

Posted 2010-05-04T21:02:58.510

Reputation: 30 733

Here "True" means "Holds in some model of ZFC"? – Louigi Addario-Berry – 2011-03-11T15:39:52.387

4@Louigi: No, "true" means "holds of the natural numbers." For example, we say that "there are infinitely many twin primes" is true if and only if there are infinitely many twin primes.

Some people will claim not to know what it means to say that there are infinitely many twin primes. But if we step next door into the number-theory classroom, then suddenly their alleged confusion disappears and they know exactly what it means to say that there are infinitely many twin primes. So all you need to do is to avoid deleting your memory as you walk back from the number theory classroom. – Timothy Chow – 2011-03-11T16:17:53.313

2If you don't know what it means to say that there are infinitely many twin primes, then you don't know what it means to say that ZFC is consistent. (Both are of the form $\forall\, n, P(n)$ for decidable $P$.) It's a respectable attitude to take that we don't really know what this means, that its truth may depend on the model, etc; but then you have to apply this across the board. (Incidentally, this is the same trap that postmodernist philosophers fall into when they try to analyse science that they don't understand.) – Toby Bartels – 2011-04-04T07:24:34.900

"Both are of the form ∀n,P(n) for decidable P." Can you elaborate further? It seems to me that there is no way to disprove "there are infinitely many twin primes" by a counterexample, whereas there is for "ZFC is consistent." – Ibrahim Tencer – 2015-03-09T16:06:33.200

@Ibrahim : You're correct. However, there are plausible $\Pi^0_1$ strengthenings of the twin prime conjecture that give computable bounds on the spacing between twin primes. – Timothy Chow – 2015-03-09T20:59:14.427

About the OP: if you are comparing ZFC natural numbers to real life as in "if you take two apples and two apples you get four apples", then arithmetic consistency is not a formal property but it is clearly not the same as consistency (unless you think the theory of real numbers can't be consistent because it also doesn't model real life natural numbers). But if you are treating it as a formal statement this requires some metatheory whose definition of the naturals may or may not reflect reality too. I guess this is what Toby meant by "you have to apply this across the board." – Ibrahim Tencer – 2015-03-09T21:57:53.137


If $f$ is (Lebesgue) integrable on $\mathbb{R}$, then $f(x) \to 0$, as $x \to \infty$. False: there exists a continuous integrable function on $\mathbb{R}$ such that $\limsup_{\infty} f = \infty$ (an exercise in Stein and Shakarchi's Real Analysis).

Slobodan Simić

Posted 2010-05-04T21:02:58.510

Reputation: 150

It's easy to prove the true statement, but I was startled at first by having it pointed out, because I recognize your false statement as something I've heard many physicists claim. Presumably they are assuming (knowingly or not) that $f$ is reasonably regular at $\pm \infty$. – Mark Meckes – 2010-06-01T16:12:50.593

I think some hold this belief by analogy with the behavior of infinite series. At least, that's the common thread I've noticed with some confused students. – Andrés E. Caicedo – 2010-06-02T01:49:58.723

5On the other hand, this is true for uniformly continuous functions (which is what most students would have in their head in the first place). – Peter Humphries – 2010-06-02T02:03:01.893

3@Peter: that probably explains the physicists as well, since I expect they are thinking of smooth functions such that $f'(x)\to 0$ as $x\to \pm \infty$, which in particular implies uniform continuity. – Mark Meckes – 2010-06-02T12:07:13.513

On the other hand, physicists have no problem with saying that $e^{ikx}$ belongs to the $L^2$ Hilbert space. Actually, they are right: (our) Lebesgue integral isn't the right instrument to formalize (their) physical intuition. – Victor Protsak – 2010-06-10T06:20:34.590

I had a quantum mechanics course last semester where they actually defined $L^2$-spaces like that, but still didnt have any problem stating $e^{ikx} \in L^2$. Physicists are weird sometimes. – Stefan Rigger – 2016-12-11T22:44:18.643


The commutator [H,K] of two subgroups H,K is the set of commutators [h,k] with h in H and k in K. (Instead, it is the group generated by those commutators. Confusingly, the convention with products HK usually goes the other way.)

In a similar vein: a 2-vector $\omega \in \bigwedge^2 V$ is always the wedge product of two 1-vectors. (Instead, this is merely an important special case of a 2-vector.) Part of the difficulty here is that the statement is true in the important three-dimensional case. Once one is aware of the Plucker embedding, this confusion goes away, but that can take a while...

Terry Tao

Posted 2010-05-04T21:02:58.510

Reputation: 53 985

3I used to have the commutator subgroup false belief, but playing around with a Rubik's cube disabused me of that notion really quickly... – Harrison Brown – 2011-05-08T11:18:57.167

2I remember teaching in a second year algebra exercise session and telling my students that "of course" the set of commutators is not necessarily a subgroup. Then they asked me for an example... And I was unable to provide an elementary one at once, the only example I could come up with in a minute was free groups, which they had never seen so I could not tell them, and it is not quite easy anyway. As it turns out, the smallest finite group for which this occurs has order > 100 (I cannot recall right now). – Arnaud Chéritat – 2015-10-18T12:08:58.710

The smallest such groups are two of order 96, as stated here.

– Rosie F – 2016-06-23T09:16:40.770

$SL_2(\mathbb R)$ is a good example. The only element that is not a commutator is $-I$. – Tom Goodwillie – 2017-07-13T15:10:55.763


Real projective space ${\mathbb{RP}}^3 = (\mathbb R^4 - 0)/\mathbb R^*$ is non-orientable.

Sam Nead

Posted 2010-05-04T21:02:58.510

Reputation: 10 750

21"Non-orientable surfaces do not embed in orientable three-manifolds." is also a classic. – Sam Nead – 2011-04-10T19:33:18.717

2"Covers of planar surfaces are planar." – Sam Nead – 2011-11-17T11:16:35.267


"A continuous image of a locally compact space is locally compact."

This is tempting because it is true without the "locally"s and it is often the case that topological properties and statements can be "localized". This came up in a problem session for my [number theory!] course this semester, and although the students were too experienced to accept it without proof, they had no alarm bells in their heads to prevent them from entertaining the possibility.

The way to quash this (as well as Andrew L.'s answer, which reminded me of this) is to realize that if it were true, every space $X$ would be locally compact, since the identity map from $X$ endowed with the discrete topology to $X$ is a continuous bijection.

Pete L. Clark

Posted 2010-05-04T21:02:58.510

Reputation: 49 252

5@Pete: related story. Is it true that a continuous image of a compact rigid space is compact? Recall that rigid spaces (in Tate's sense) only have a Grothendieck topology on them, not a topology, so "compact" means that every admissible cover has an admissible finite subcover. The problem with the usual proof is that you admissibly cover the target, pull it back to the source, write down a finite admissible subcover, push it forward, and it might not be admissible! Ofer Gabber once presented me with a counterexample to the assertion written on the back of a napkin :-) – Kevin Buzzard – 2010-05-04T22:13:40.800

4"Recall" -- what, from your course in 2002(ish)? No problem, this is one of the few things I remember lo these many years later. I particularly recall your intuition for Tate's Grothendieck topology as requiring the open sets to be "big and chunky". – Pete L. Clark – 2010-05-04T22:20:18.527

2That there are two different notions that are called "locally compact" does not help. – Alfonso Gracia-Saz – 2010-05-07T21:21:25.140


I just finished quadratic congruences in my number theory class. I am not any more surprised to see how strong is students' belief in the fact that $x^2\equiv a\pmod{m}$ has at most 2 solutions. Even after you discuss an example $x^2\equiv1\pmod{143}$ (with solutions $x\equiv\pm1,\pm12\pmod{143}$) in details.

And, of course, a lot of wrong beliefs in real analysis. Like an infinitely differentiable function, say in a neighbourhood of origin, must be analytic at the origin.

Wadim Zudilin

Posted 2010-05-04T21:02:58.510

Reputation: 10 081

1I think you mean that they think $x^2 \cong a \pmod m$ can only have 2 solutions. This is the case over the reals, which is probably where the students get it from. – Michael Lugo – 2010-05-05T00:33:16.053

1@Michael: It might have 0 and 1 solutions as well (for example, when $m$ is prime and $a$ is a quadratic nonresidue or 0 respectively). The "real" analogues are $x^2=-1$ and $x^2=0$. – Wadim Zudilin – 2010-05-05T01:05:59.917

2Michael's comment is because you have said that students believe that $x^2 \equiv a \pmod m$ may have more than $2$ solutions. But as you go on to indicate, this is a correct belief. So there must be a typo in your answer. – Pete L. Clark – 2010-05-05T02:19:53.093

Well.. It is true when 143 is prime. – VictorZurkowski – 2017-04-12T14:32:30.497

Perhaps solving $x^2 \equiv 1 \pmod8$ is more memorable than solving $x^2 \equiv 1 \pmod{143}$? – LSpice – 2018-01-05T21:05:57.350


$\pi$ is equal to 22/7.

This was touched upon in the comments to a totally unrelated answer but I think this false belief is important enough to warrant its own answer (and as far as I could tell it does not have one yet, my apologies if I overlooked one.)

Of course, it's unlikely anyone on this site believes this, or ever believed it, which is why I think it's important to insist on this: it does not really resonate with us, we are unlikely to warn students against it, yet we probably see in front of us many students who have that false belief and then will move on to spread it around.

A Piece of Evidence

Let me offer as evidence this gem taken off the comments section of an unrelated (but quite thought-provoking) article on Psychology Today, of all places! When Less is More: The Case for Teaching Less Math in Schools (The title is a misnomer, it's a case for starting math later, but I think that with such a scheme you should be able to teach more math overall; anyway, read it for yourselves.)

Some years ago, my (now ex-) wife was involved in a "trivia night" fundraiser at her elementary school, and they wanted me on their "teacher team" to round out their knowledge. They had almost everything covered except some technology-related topics and I was an IT guy. In round four, my moment to shine arrived, as the category was "Math & Science" and one of the questions was, "give the first five digits of pi." I quickly said, "3.1415." The 9 teachers at the table ignored me and wrote down "22/7" on scrap paper and began to divide it out. I observed this quietly at first, assuming that 22/7ths gave the right answer for the first 5 digits, but it doesn't. It gives something like 3.1427. I said, "Whoops, that won't work." They ignored me and consulted among themselves, concluding that they had all done the division properly on 22/7ths out to five digits. I said, "That's not right, it's 3.1415." [...]

I'm cutting it off here because it's a long story: hilarity ensues when the non-teacher at the table stands up for the truth (when he finds out that the decimals of 22/7 were the expected answer!) The final decision of the judges:

"We've got a correction on the 'pi' question, apparently there's been confusion, but we will now be accepting 3.1415 as a correct answer as well" [as 3.1427].

The Moral of the Story

I used to dismiss out of hand this kind of confusion: who could be dumb enough to believe that $\pi$ is 22/7? (Many people apparently: in the portion of the story I cut was another gem - "I'm sorry, but I'm a civil engineer, and math is my job. Pi is 22/7ths.")

Now, I treat this very seriously, and depending on where you live, you should too. Damage wrought during the influential early years is very hard to undo, so that the contradictory facts "$\pi$ is irrational" and "$\pi$=22/7" can coexist in an undergraduate's mind. And when that person leaves school, guess which of the two beliefs will get discarded: the one implanted since childhood, or the one involving a notion (rational numbers) which is already getting fuzzy in the person's brain? I'm afraid it's no contest there, unless this confusion has been specifically addressed.

So if you have any future teachers in your classes (and even if you don't, cf. the civil engineer above), consider addressing this false belief at some point.

Thierry Zell

Posted 2010-05-04T21:02:58.510

Reputation: 3 516

6You know what the most funny part is? They can't divide, because actually 22/7 is between 3.1428 and 3.1429, but this way they got something closer to the real answer. – Zsbán Ambrus – 2010-11-27T19:36:13.207

@Zsban: the guy does write "something like 3.1427" so I'm guessing he got the digits from memory, in which case he was really close. – Thierry Zell – 2010-11-28T02:21:46.667

My father-in-law, an engineer, thought that $\pi$ was $22/7$ until I explained to him otherwise. – Ravi Boppana – 2010-12-01T16:46:04.000

6Well, doesn't the Bible say that π=3 ? – ACL – 2010-12-01T22:52:25.067

2ACL: I don't have the passage handy now, but as I recall, it does not outright say perimeter over diameter equals three; you can make the case that it's a matter of overenthusiastic rounding down of the value of the perimeter. It is a bit sloppy though, even for that time period... – Thierry Zell – 2010-12-02T02:00:43.377

@ACL: Oh, bonjour Antoine et bienvenue sur MO! – Thierry Zell – 2010-12-02T03:09:15.947

5This one is depressing... – Andrés E. Caicedo – 2010-12-07T22:26:10.473


This is from the AFP report on the recent computation of the first $5\times 10^{12}$ (decimal) digits of $\pi$: "Pi, the ratio of a circle's circumference to its diameter, starts with 3.14159 in a string whose digits are believed to never repeat or end." http://www.google.com/hostednews/afp/article/ALeqM5i8R8sLPPRJvax-BjAtH0-lsV3kGw

– Andrés E. Caicedo – 2010-12-07T22:29:12.380

3@Andres: Well, at least these folks believe something that's true, unlike the ones in Thierry's answer. – Andreas Blass – 2010-12-07T22:43:44.357

This false belief has nevertheless a noble origin, I think: Archimedes bound, $3+10/71<\pi<3+1/7.$ – Pietro Majer – 2011-01-12T11:33:30.030

2I agree with the Psychology Today article. The practice of requiring everyone to study math has resulted in most people being taught lies. – Michael Hardy – 2011-02-06T20:51:25.053

@Michael Hardy: care to elaborate further? $$*$$ In (at least) one of his books, Ian Stewart has a liar-to-children character. This is the teacher, and his role is revered, with the understanding that in order to teach, one may have to simplify at first, and only progressively paint a more complicated picture that fits reality better. This is in contrast to the liar-to-adult (politician) who is universally reviled. Saying that pi is 22/7 is not a good lie to tell children, but saying that the probability of rolling a 4 on a die is 1/6 might be. – Thierry Zell – 2011-02-06T23:23:50.443

So who first observed that $\int_0^1 (x-x^2)^4 dx/(x^2+1) = (22/7) - \pi$? – Noam D. Elkies – 2011-06-05T16:00:18.800

Upper bound: This is problem A-1 on the 29th Putnam Exam (1968), whose Questions Committee consisted of N.D.Kazarinoff, Leo Moser, and Albert Wilansky according to the report in the American Math. Monthly (Vol.76 (1969) #8, 909-915). The solution there says only "The standard approach, from elementary calculus, applies. By division, rewrite the integrand as a polynomial plus a rational function with numerator of degree less than 2. The solution follows easily". – Noam D. Elkies – 2011-06-05T20:58:24.060

If you believe in sympy:
In [7]: integrate( (x4 * (1-x)4)/(1+x*x), (x,0,1))

Out[7]: -π + 22/7

and then point out the integrand is continuous and positive. – kjetil b halvorsen – 2013-01-14T22:11:45.100

2@ACL What it says on the topic is (1Kings 7:23): "And he made a molten sea, ten cubits from the one brim to the other: it was round all about, and his height was five cubits: and a line of thirty cubits did compass it round about." Well, one may read this as "$\pi$ equals 3" (probably, any other numerical data in such a passage would at least imply that $\pi$ is rational.) – Fedor Petrov – 2016-01-02T21:16:03.723

I believed that $\pi$ was $22/7$ because in 9th grade my textbook said so. I know better now, but I'm not sure of any way to convincingly argue without calculus or authority that it isn't. – Kevin O'Bryant – 2016-09-05T04:03:25.557


All higher homotopy groups of spheres are zero.

Proof: The higher homology groups of spheres are zero, and the higher homotopy groups are abelian, and since homology groups are abelianizations of homotopy groups the higher homotopy groups are alse zero.

This misconception is also made more difficult by the fact that even the simple counterexamples can't be drawn easily.


Posted 2010-05-04T21:02:58.510

Reputation: 451

1For anyone who knows the definition of the projective line, the equivalence classes on the 3-sphere in C^2 defined by complex line sections through the origin, has equivalence space CP^1 ≈ S^2, i.e. the Hopf map. I believe I thought of this as a student, but probably did not show it is non trivial. – roy smith – 2011-04-14T18:51:36.410

3The Hopf map is pretty elementary... – Andy Putman – 2010-05-05T00:47:37.297

18I agree that it's elementary, but it's not a particularly easy example to come up with on your own. – Inna – 2010-05-05T02:19:17.787

Yes, contradicting Hurewicz heorem ... – Qfwfq – 2010-05-05T12:13:21.493

This seems to be a constellation of false ideas (some of which I held until corrected by my advisor): $\pi_n(X)$ tends to be nonabelian for all $n$. (In fact $\pi_n(X)$ is abelian for $n \geq 2$. See https://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument where it is related that Alexandroff and Hopf were surprised by this.) $H_n(X)$ is (isomorphic to) the abelianization of $\pi_n(X)$ for all $n$. (True for $n \neq 2$. False for $n=2$ unless $X$ is $1$-connected. The Hurewicz theorem gives a homomorphism, $\pi_2 \rightarrow H_2$ when $X$ is not $1$-connected.)

– Eric Towers – 2017-08-13T02:52:58.000

Actually, I fumbled the last parenthetical. Relative to an $n-1$-connected subspace, $A \subset X$, Hurewicz gives an isomorphism between the abelianization of $\pi_n(X,A)$ and $H_n(X,A)$. For $n \geq 3$, this is trivial, essentially because the $\pi_n$ are already abelian. I fumbled the $n=2$ case. Relative to Inna's Answer, this only tells us about the relationship between the first nonzero homotopy group and its corresponding homology group. It give us no information about the higher pairings. – Eric Towers – 2017-08-13T04:19:51.550


This is a bit specialized, but a common misconception in low-dimensional topology (particularly in knot theory) is that any change of basis in homology is realized by a diffeomorphism, hence (for a surface) by an action of a mapping class. I think this is exactly the type of false belief being described (I falsely believed it for a long time myself).

Common misconception: Let F be a genus 2g surface, and let $b_1,\ldots,b_{2g}$ be a primitive basis for $H_1(F)$, represented as embedded curves in F. Any change of basis for $H_1(F)$ is realized by an action of the mapping class group of F on the embedded curves.

This is rubbish- the action of the mapping class group on homology is by $Sp_{2g}(\mathbb{Z}))$, which for $g>1$ is a proper subgroup of $GL_{2g}(\mathbb{Z})$, the group of base-changes of $H_1(F)$.
As an example of what you can't do with a diffeomorphism of a surface, consider a disc with 4 bands A,B,C,D attached, so the order of the end sements is $A^+B^-A^-B^+C^+D^-C^-D^+$, together forming a surface. A basis a,b,c,d for $H_1(F)$ is given by this picture as 4 loops going through the cores of the bands A,B,C,D correspondingly. You can add a to b, b to c, c to d, or d to a by diffeomorphism of F (sliding adjacent bands over one another). However, although you can add a to c algebraically, because bands A and C are "not adjacent in F", there is no corresponding diffeomorphism of $F$.
One place this mistake manifests itself (cranking up the level of terminilogy for a second) is in thinking that unimodular congruence of a Seifert matrix corresponds to ambient isotopy of a Seifert surface.
A related common mistake (closely related to this question):

Common misconception: Any homology class is represented as a submanifold. Maybe even as an embedded submanifold.

Daniel Moskovich

Posted 2010-05-04T21:02:58.510

Reputation: 12 674

2Is the first common misconception actually common? I have never met anyone who had it (perhaps I hang out with the wrong crowd...) – Igor Rivin – 2011-01-28T22:51:52.043

Igor, I guess your crowd knows about intersection pairings. – roy smith – 2011-04-14T18:59:50.467

Could you give a counterexample for the 2nd misconception? – Michael – 2013-10-31T08:04:42.207


A false belief which I meet not infrequently this time of year while marking exams is the following:

The exponential map is surjective for a connected Lie group.

This is true for compact Lie groups, but certainly false in general. A (finite-dimensional) connected Lie group is generated by the image of the exponential map, but already $SL(2,\mathbb{R})$ shows that there are elements which are not in the image of the exponential map.

Interestingly, for a connected real Lie group, every element can be written as the product of at most two exponentials.

José Figueroa-O'Farrill

Posted 2010-05-04T21:02:58.510

Reputation: 24 255

Well, exponential function isn't surjective either, is it? More seriously, not only does the exp map fail to be surjective, it also fails to be a local diffeomorphism! But I wonder what beliefs like that say about us. Certainly, unexpected things happen when you push familiar notions into an unfamiliar terrain. Isn't it our job as teachers to make the students perfectly aware of what properties do and do not continue to hold? Where is that intuition supposed to come from? I once taught a Lie groups course, and the book I used (Rossmann) made a big deal about matrix exponentials being unusual. – Victor Protsak – 2010-06-10T06:36:48.010

1But the exponential function, thought of as the exponential map from the one-dimensional Lie algebra $\mathbb{R}$ to the Lie group $\mathbb{R}^+$ (under multiplication) is surjective.

I too teach this course based on Wulf Rossmann's book and I think that I do emphasise, if not perhaps all the unusualness of the exponential map, certainly the fact that it may fail to be surjective on the identity component. We even do the case of $SL(2,\mathbb{R})$ is some detail as an application of the formula for the exponential of a $2\times 2$ matrix. – José Figueroa-O'Farrill – 2010-06-10T08:12:52.563


Let $A,B$ be Hermitian matrices. If $0_n\le A\le B$, then $A^2\le B^2$.

False, but subtle! Loewner's theory characterizes those numerical functions $f:[0,\infty)\rightarrow {\mathbb R}$ such that $0_n\le A\le B$ implies $f(A)\le f(B)$ (operator monotone functions). These are the traces over $[0,\infty)$ of holomorphic functions mapping the Poincaré half-plane ${\mathcal H}$ into itself, and of course real on $[0,\infty)$. Thus the square root is operator monotone: $$(0_n\le A\le B)\Longrightarrow(\sqrt A\le\sqrt B),$$ but the square map is not. Counter-example: $$A=\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} 2 & 1 \\\\ 1 & 1 \end{pmatrix}$$

Denis Serre

Posted 2010-05-04T21:02:58.510

Reputation: 26 530

I hope it is true when $A,B$ commute. ;) – Martin Brandenburg – 2011-04-12T08:49:24.710

I have heard that Ed Witten got this one wrong in a paper. – J Tyson – 2015-10-25T22:13:04.537


The quotient $G/Z(G)$ of a group by its center is centerless. I definitely thought this until it was pointed out to me in a Lie theory textbook that this wasn't true in general, but is true for (edit: connected) Lie groups with discrete center. (It is also true if $G$ is perfect by Grun's lemma.)

Qiaochu Yuan

Posted 2010-05-04T21:02:58.510

Reputation: 73 568

9For a counter-example, just consider a non-abelian $p$-group $G$. Because $G/Z(G)$ is still a $p$-group, its center is non trivial (classical). – Denis Serre – 2010-10-20T10:51:15.980

Indeed. Though in your second sentence I guess your Lie group has finitely may components? For a concrete example, Heisenberg group (over ${\mathbb R}$ or ${\mathbb Z}$) will be abelian modulo its centre... – Yemon Choi – 2010-07-10T03:47:16.740

13If that were true, then the class of Abelian groups and nilpotent groups would coincide! – Abhishek Parab – 2010-07-10T03:51:01.477

@Yemon: oops. I wanted "connected." – Qiaochu Yuan – 2010-07-10T04:07:01.317


$0^0$ is undefined.

EDIT: People write things like $\sum_{k=0}^\infty x^k$ all day, but somehow $x=k=0$ is still scary when written as $0^0$.


Posted 2010-05-04T21:02:58.510

Reputation: 1

1This appears in (pretty much) every high-school algebra text in the U.S., even elementary ones that only consider rational exponents. Yes, it's a convention, but they've clearly chosen the wrong one for the context. – Toby Bartels – 2011-04-04T07:45:49.820

6Are you saying that's a false belief?? – Kevin Buzzard – 2010-05-04T22:11:09.663

13This isn't a false belief so much as a matter of convention. I think the OP is after statements which are believable and provably false. – Qiaochu Yuan – 2010-05-04T22:17:48.583

3@Kevin: yes. @Qiaochu: it is a belief incompatible with "$\sum_{k=0}^\infty x^k$ is defined for $x=0$" which is usually considered true. – B C – 2010-05-04T22:26:44.580

5Yes, but the opposite belief is incompatible with the statement that a^b = e^{b ln a}. In other words, this is a matter of convention (or definition, if you prefer) and it differs based on context. It's not a false mathematical statement in the usual sense. – Qiaochu Yuan – 2010-05-04T22:55:39.673

430^0 = 1, because there is one function from the empty set to the empty set – Steven Gubkin – 2010-05-04T23:04:09.340

25Some would say ... $0^0=1$ where the exponent is the integer $0$, but $0^0$ is undefined where the exponent is the real number $0$. Does this fit all of the comments so far? – Gerald Edgar – 2010-05-05T00:15:29.577

4@Qiaochu: No, this is not a matter of convention. If you claim that 0^0 is undefined, then every algebra textbook contains a mistake on almost every page, because you cannot write polynomials or power series in the form ∑_k a_k x^k. – Dmitri Pavlov – 2010-05-05T00:52:58.760

8@Qiaochu: More precisely, the notation a^b denotes two completely different notions. The first notion requires that a is an element of some monoid and b∈N={0,1,2,…}. The second notion requires that a∈C and b belongs to the universal cover of C \ {0}. Then a^b = exp(b * Log(a)), where Log denotes the natural logarithm defined as a function on the universal cover of C \ {0}. Accidentally, these two functions coincide when a∈C and b∈N \ {0}. – Dmitri Pavlov – 2010-05-05T00:53:11.473

5The limit of x^y as x, y -> 0 does not exist. The power series notation works because it is clear in that context that you are fixing y and varying x, so it is obvious what the convention should be. – Charles Staats – 2010-05-05T03:40:17.337

22@Dmitri Pavlov: I disagree. A polynomial is a purely algebraic object. The notation $t^0$ is just shorthand for a certain polynomial, otherwise called $1$. (If you identify $R[t]$ with the set of finitely nonzero functions from $\mathbb{N}$ to $R$, it is the function which maps $0$ to $1 \in R$ and everything else to zero.) Evaluation of a polynomial at an element of $R$ is a certain ring homomorphism $R[t] \rightarrow R$ which in particular sends the polynomial $1$ to the element $1$. We never "evaluate" $0^0$ in any sense, because indeed it has no independent algebraic definition. – Pete L. Clark – 2010-05-05T06:31:23.193

3@Dmitri Pavlov: It denotes a third notion as well: if A and B are objects in a cartesian/monoidal closed category, then A^B is another notation for the mapping-object [B,A]. With the category FinSet (finite sets; functions) this coincides with (or more precisely, decategorifies to) the "power" definition in the monoid N. In this case, even more than the monoid case, it's completely clear that 0^0=1, as Steven Gubkin points out. – Peter LeFanu Lumsdaine – 2010-05-05T14:39:09.663

3C.S. Peirce noted that $p^q$ for $p, q \in \lbrace 0, 1 \rbrace$ under the interpretation of $0$ as false and $1$ as true represents the proposition $p \Leftarrow q$, in other words, $p$ if $q$. In particular, $p^0 = 1$. – Jon Awbrey – 2010-05-28T02:34:55.220

5It's also interesting to note that $\lim_{x \to 0^+} x^x=1$. – David Corwin – 2010-07-07T23:20:32.497


The standard projection map in a first course in topology is open. How could it not be closed? I always forget the standard homework exercise in which people first try to use this non-fact.


Posted 2010-05-04T21:02:58.510

Reputation: 540

this is another of the standard algebraic geometry examples from the "red book", whereby one shows the affine open set f(x) ≠ 0 is isomorphic to the closed set yf(x) = 1, in one dimension higher ambient space. – roy smith – 2011-04-14T18:45:50.320

13The standard counter-example is $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, with the closed hyperbola $xy = 1$ mapping to the open set $x \neq 0$. I remember making this mistake, but not what problem prompted me to do so. – David E Speyer – 2010-05-05T12:44:53.713

@roysmith, I remember how puzzled I was when first learning about algebraic groups to be forced to think of $\mathrm{GL}_n$ as a closed subvariety of an affine space, when it so manifestly wants to be open …. – LSpice – 2018-01-05T21:07:09.173


If a topological space has an open cover by Hausdorff spaces, it is Hausdorff.


Posted 2010-05-04T21:02:58.510

Reputation: 1

1of course all algebraic geometers have some picture such as that on pages 34 of Eisenbud-Harris, or that on p. 37 of Mumford's (new) red book, tattooed on their eyeballs as a reminder of the difference between a scheme and a separated scheme. – roy smith – 2011-04-14T18:16:35.047


a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is $n$ times differentiable if and only if for each real $x_0$ it may be approximated near $x_0$ by a polynomial of degree at most $n$ with remainder $o((x-x_0)^n)$

Fedor Petrov

Posted 2010-05-04T21:02:58.510

Reputation: 39 771

3+1: I posted an answer to a question on math.SE using this "characterization" of differentiability. Didier Piau pointed out that it was completely wrong. It is true for $n = 1$, so even now I find it somewhat weird that it doesn't hold for higher derivatives... – Pete L. Clark – 2011-04-26T15:03:17.060

2It becomes true, and a deep fact, in the $C^n$ version: the function $f$ is $C^n$ iff it has a n-order polynomial expansion at any $x_0$, with coefficients depending continuously on $x_0$. – Pietro Majer – 2012-01-10T13:48:58.250

2Pietro, what do you mean exactly? "There exist continuous in $x_0$ functions $h_0(x_0)$, $h_1(x_0)$, $\dots$, $h_n(x_0)$ such that for any fixed $x_0$ one has $f(x_0+t)=h_0(x_0)+h_1(x_0)t+\dots+t^n h_n(x_0)+o(t^n)$ when $t$ goes to 0." Or something different? – Fedor Petrov – 2012-01-11T21:59:17.800

@Pete: Do you mean the lower derivatives may be wild? Do you have a good counter-example? – Kerry – 2012-07-30T06:05:00.933


I'm not sure how common this is, but it confused me for years. Let $f : \mathbb{C} \to \mathbb{C}$ be an analytic function and $\gamma$ a path in $\mathbb{C}$. In your first class in complex analysis, you define the integral $\int_{\gamma} f(z) dz$.

Now let $a(x,y) dx + b(x,y) dy$ be a $1$-form on $\mathbb{R}^2$ and let $\gamma$ be a path in $\mathbb{R}^2$. In your first class on differential geometry, you define the integral $\int_{\gamma} a(x,y) dx + b(x,y) dy$.

It took me at least three years after I had taken both classes to realize that these notations are consistent. Until then, I thought there was a "path integral in the sense of complex analysis", and I wasn't sure if it obeyed the same rules as the path integral from differential geometry. (By way of analogy, although I wasn't thinking this clearly, the integral $\int \sqrt{dx^2 + dy^2}$, which computes arc length, is NOT the integral of a $1$-form, and I thought complex integrals were something like this.)

For the record, I'll spell out the relation between these notions. Let $f(x+iy) = u(x,y) + i v(x,y)$. Then $$\int_{\gamma} f(z) dz = \int_{\gamma} \left( u(x,y) dx - v(x,y) dy \right) + i \int_{\gamma} \left( u(x,y) dy + v(x,y) dx \right)$$ The right hand side should be thought of as multiplying out $\int_{\gamma} (u(x,y) + i v(x,y)) (dx + i dy)$, a notion which can be made rigorous.

David E Speyer

Posted 2010-05-04T21:02:58.510

Reputation: 98 510

2I think it is customary to say "contour integral" for the complex analysis gadget, "line integral" for the multivariable calculus gadget, and "path integral" for a (not necessarily rigorously defined) integral over a space of fields. – S. Carnahan – 2011-01-13T02:47:09.090


As a teaching assistant in an elementary number theory course, I've seen the following quite often :

If $a$ divides $bc$ and $a$ does not divide $b$, then $a$ divides $c$.

That's of course true if $a$ is prime, but people seem to forget that hypothesis.


Posted 2010-05-04T21:02:58.510

Reputation: 814

i was never able to eradicate this mistake, even by my best students, in spite of naming it the prime divisibility property. – roy smith – 2011-05-09T02:15:29.270

12@MartinBrandenburg : What does it mean for an assumption to be incorrect or correct? – Kalim – 2014-02-07T18:40:18.180

Easy counterexample: $6 = 2 \times 3$ but $6$ does not divide $2$ or $3$... – Mehrdad – 2017-12-28T09:41:51.993

8the correct assumption is $gcd(a,b)=1$. – Martin Brandenburg – 2010-10-05T08:23:28.053


Any subgroup of the direct product $G \times H$ of two groups is of the form $A \times B$, where $A$ is a subgroup of $G$ and $B$ is a subgroup of $H$.


Posted 2010-05-04T21:02:58.510

Reputation: 2 414

What? Is it false? Could you please explain it? – Bumblebee – 2016-03-03T09:19:14.290

3@Nilan: For $G=H$, consider the diagonal subgroup ${ (g,g) : g \in G}$. It's not of the form above unless $G$ is the trivial group. – Mark – 2016-03-03T10:32:41.337


I'm not sure that anyone holds this as a conscious belief but I have seen a number of students, asked to check that a linear map $\mathbb{R}^k \to \mathbb{R}^{\ell}$ is injective, just check that each of the $k$ basis elements has nonzero image.

David E Speyer

Posted 2010-05-04T21:02:58.510

Reputation: 98 510

16Higher-level version: $n$ vectors are linearly independent iff no two are proportional. I've seen applied mathematicians do that. – darij grinberg – 2011-04-10T18:45:17.103


After learning that the Witt vectors of a finite field of size $p^n$ is the ring of integers of the unramified extension of ${\mathbf Q}_p$ of degree $n$, I think lots of people then think that the Witt vectors of $\overline{\mathbf F}_p$ (the algebraic closure of ${\mathbf F}_p$) is the ring of integers of the maximal unramified extension of ${\mathbf Q}_p$. It isn't: the integers of the maximal unramified extension is the union of the Witt vectors of the finite fields of $p$-power size whereas the Witt vectors of $\overline{\mathbf F}_p$ is the $p$-adic completion of the integers of the maximal unramified extension; the distinction turns on being able to write Witt vectors over $\overline{\mathbf F}_p$ as series with coefficients that are prime-to-$p$ roots of unity of increasingly large degree instead of having bounded degree.

I was at a conference last fall where a famous mathematician was confused by this point, although to be fair he really never worked seriously with Witt vectors before.


Posted 2010-05-04T21:02:58.510

Reputation: 27 105

1If you're going to hold a false belief, this is a good one to have. While the maximal unramified extension of $\mathbb{Q}_p$ is not the fraction field of $W(\overline{\mathbb{F}_p})$ -- the latter is complete, the former isn't -- one is a subfield of the other and the two fields have very similar properties. One way to express/justify this is by saying that the inclusion map is an elementary embedding in the sense of model theory. For arithmetic geometers, this is related to work of Greenberg on rational points over Henselian fields. – Pete L. Clark – 2010-05-05T02:25:11.330

Pete: Witt vectors are confusing enough when you're trying to learn about them that I think this is a false belief that is important to clear up if, say, your thesis depends on understanding Witt vectors. – KConrad – 2010-05-05T04:13:23.597

1@K: Well, sure, believing false things is never desirable. All I'm saying is that there are some things working underneath the surface to prevent this particular false belief from really screwing you up. I seem to remember once writing something like "where $\mathbb{Q}_{p^{\infty}}$ is, according to the reader's preference, either the maximal unramified extension of $\mathbb{Q}_p$ or its completion." Not that this is especially good mathematical writing, but the point was that it manifestly didn't matter which. – Pete L. Clark – 2010-05-05T04:57:20.737

P.S.: $W(\overline{\mathbb{F}_p})$ does appear in my thesis...I hope I got it right. :) – Pete L. Clark – 2010-05-05T06:39:33.460

5Pete: It can screw you up very badly. For example, F-isocrystals are very different over the maximal unramified extension of Q_p and its completion. – JS Milne – 2010-05-08T16:44:59.530

@Prof. Milne: I take your point. Thanks. – Pete L. Clark – 2010-05-08T19:24:58.593

A marginally constructive remark: the ring of integers in $\mathbb Q_p^{nr}$ is the strict henselization of $\mathbb Z_p=W(\mathbb F_p)$. – inkspot – 2010-08-26T15:48:37.403


False belief: Every commuting pair of diagonalizable elements of $PSL(2,\mathbb{C})$ are simultaneously diagonalizable. The truth: I suppose not many people have thought about it, but it surprised me. Look at $$\left(\matrix{i& 0 \cr 0 & -i\cr } \right), \left(\matrix{0& i \cr i & 0\cr } \right).$$

Ben McKay

Posted 2010-05-04T21:02:58.510

Reputation: 11 798

5To me, it is marvellous that the failure of this fact (as opposed to the truth of the corresponding fact for $\operatorname{SL}(2, \mathbb C)$) is a matter of topology; that is, from the point of view of algebraic groups, it comes from the fact that $\operatorname{SL}(2, \mathbb C)$ is simply connected, whereas $\operatorname{PSL}(2, \mathbb C)$ (which I had rather call $\operatorname{PGL}(2, \mathbb C)$) is not (it is at the opposite extreme---`adjoint'). – LSpice – 2013-12-12T23:09:46.460


In measure-theoretic probability, I think there is sometimes an idea among beginners that independent random variables $X,Y$ should be thought of as having "disjoint support" as measurable functions on the underlying probability space $\Omega$. Of course this is the opposite of the truth.

I think this may come from thinking of measure theory as generalizing freshman calculus, so that one's favorite measure space is something like $[0,1]$ with Lebesgue measure. This is technically a probability space, but a really inconvenient one for actually doing probability (where you want to have lots of random variables with some amount of independence).

Nate Eldredge

Posted 2010-05-04T21:02:58.510

Reputation: 17 394

1This disjoint support misconception reinforces the incorrect idea that pairwise independent implies independent. – Douglas Zare – 2010-10-20T18:47:25.087

1A student this last semester made precisely this mistake, and it was a labor of three people to convince him otherwise. – Andrés E. Caicedo – 2010-05-17T00:28:09.827


"Euclid's proof of the infinitude of primes was by contradiction."

That is a very widespread false belief.

"Prime Simplicity", Mathematical Intelligencer, volume 31, number 4, pages 44--52, by me and Catherine Woodgold, debunks it. The proof that Euclid actually wrote is simpler and better than the proof by contradiction often attributed to him.

Michael Hardy

Posted 2010-05-04T21:02:58.510

Reputation: 5 412

1Since Euclid omits to check that there exists at least one prime, his induction must begin with the case n = 0, hence his argument seems to require the fact that the product of an empty set of primes equals 1. – roy smith – 2011-05-09T02:27:47.077

2@roy: if there are no primes, there are finitely many a fortiori. So there's nothing to check there. – Toink – 2013-03-18T00:04:44.127

@roysmith : Euclid didn't even consider $1$ to be a number. $\qquad$ – Michael Hardy – 2016-03-10T05:15:47.307

1And you'd be surprised how many quite knowledgable PHD's spend decades repeating this mistake to thier students,Micheal. – The Mathemagician – 2010-06-07T00:07:24.980

5Actually, if you read our paper on this, you'll find that I won't be surprised at all.

(BTW, my first name is spelled in the usual way, not the way you spelled it.) – Michael Hardy – 2010-06-07T03:28:12.870

wow! I tend to take historical statements in math books/papers with a grain of salt, but this makes me think more salt might be required... – Rob Harron – 2010-06-12T01:38:55.270

1So what was Euler's true original proof, then? – BlueRaja – 2010-07-07T21:36:01.910

3@ BlueRaja: I'm assuming "Euler" is a typo and you meant Euclid.

Euclid said if you take any arbitrary finite set of prime numbers, then multiply them and add 1, and factor the result into primes, you get only new primes not already in the finite set you started with.

The proof that they're not in that set is indeed by contradiction. But the proof as a whole is not, since it doesn't assume only finitely many primes exist. – Michael Hardy – 2010-07-07T21:55:14.907

Yes, Euclid sorry :) – BlueRaja – 2010-07-08T20:29:08.690

4Note indeed the original Euclid's statement: Prime numbers are more than any previously assigned finite collection of them (my translation). This reflects a remarkable maturity and consciousness, if we think that mathematicians started speaking of infinite sets a long time before a well founded theory was settled and paradoxes were solved. Euclid's original proof in my opinion is a model of precision and clearness. It starts: Take e.g. three of them, A, B and Γ . He takes three prime numbers as the first reasonably representative case to get the general construction. – Pietro Majer – 2010-07-20T14:51:23.310

1Actually I think the use of three letters was just a notational device. He clearly meant an arbitrary finite set of prime numbers (if he hadn't had that in mind, he couldn't have written that particular proof). – Michael Hardy – 2010-07-20T22:43:38.777


In descriptive set theory, we study properties of Polish spaces, typically not considered as topological spaces but rather we equip them with their "Borel structure", i.e., the collection of their Borel sets. Any two uncountable standard Borel Polish spaces are isomorphic, and the isomorphism map can be taken to be Borel. In practice, this means that for most properties we study it is irrelevant what specific Polish space we use as underlying "ambient space", it may be ${\mathbb R}$, or ${\mathbb N}^{\mathbb N}$, or ${\mathcal l}^2$, etc, and we tend to think of all of them as "the reals".

In Lebesgue Sur les fonctions representables analytiquement, J. de math. pures et appl. (1905), Lebesgue makes the mistake of thinking that projections of Borel subsets of the plane ${\mathbb R}^2$ are Borel. In a sense, this mistake created descriptive set theory.

Now we know, for example, that in ${\mathbb N}^{\mathbb N}$, projections of closed sets need not be Borel. Since we usually call reals the members of ${\mathbb N}^{\mathbb N}$,

it is not uncommon to think that projections of closed subsets of ${\mathbb R}^2$ are not necessarily Borel.

This is false. Note that closed sets are countable union of compact sets, so their projections are $F_\sigma$. The actual results in ${\mathbb R}$ are as follows: Recall that the analytic sets are (the empty set and) the sets that are images of Borel subsets of $\mathbb R$ by Borel measurable functions $f:\mathbb R\to\mathbb R$.

  • A set is Borel iff it and its complement are analytic.

  • A set is analytic iff it is the projection of the complement of the projection of a closed subset of ${\mathbb R}^3$.

  • A set is analytic iff it is the projection of a $G_\delta$ subset of $\mathbb R^2$.

  • There is a continuous $g:\mathbb R\to\mathbb R$ such that a set is analytic iff it is $g(A)$ for some $G_\delta$ set $A$.

  • A set if analytic iff it is $f(\mathbb R\setminus\mathbb Q)$ for some continuous $f:\mathbb R\setminus\mathbb Q\to\mathbb R$. (Note that if $f$ is actually continuous on $\mathbb R$, then $f(\mathbb R\setminus\mathbb Q)$ is Borel.)

(See also here.)

Andrés E. Caicedo

Posted 2010-05-04T21:02:58.510

Reputation: 24 270


"Suppose that two features $[x,y]$ from a population $P$ are positively correlated, and we divide $P$ into two subclasses $P_1$, $P_2$. Then, it cannot happen that the respective features ( $[x_1,y1]$ and $[x_2,y_2]$) are negatively correlated in both subclasses

Or more succintly:

"Mixing preserves the correlation sign."

This seems very plausible - almost obvious. But it's false - see Simpon's paradox


Posted 2010-05-04T21:02:58.510

Reputation: 216

This seems like much more than a mathematical mistake - really it's a cognitive bias which incidentally can be turned into a mathematically false statement. – Qiaochu Yuan – 2011-05-09T22:53:45.667

There were many cases of this error made by reporters and even courts. – Michael – 2013-12-03T01:18:59.027


In group theory, if $G_1 \cong G_2$ and $H_1 \cong H_2$, then

$G_1 / H_1 \cong G_2 / H_2$.

For example, $\mathbb{Z} / 2\mathbb{Z} \not \cong \mathbb{Z} / \mathbb{Z}$. The point is that the inclusion of $H_j$ into $G_j$ is needed in order to define the quotient.

André Caldas

Posted 2010-05-04T21:02:58.510

Reputation: 317


A common misbelief for the exponential of matrices is $AB=BA \Leftrightarrow \exp(A)\exp(B) = \exp(A+B)$. While the one direction is of course correct: $AB=BA \Rightarrow \exp(A)\exp(B) = \exp(A+B)$, the other direction is not correct, as the following example shows: $A=\begin{pmatrix} 0 & 1 \\ 0 & 2\pi i\end{pmatrix}, B=\begin{pmatrix} 2 \pi i & 0 \\ 0 & -2\pi i\end{pmatrix} $ with $AB \neq BA \text{ and} \exp(A)=\exp(B) = \exp(A+B) = 1$.

Andreas Rüdinger

Posted 2010-05-04T21:02:58.510

Reputation: 2 301

5Related to the mistake mentionned by David, the fact that the solution of a vector ODE $x'(t)=A(t)x(t)$ should be $$\left(\exp\int_0^tA(s)ds\right)x(0).$$ – Denis Serre – 2010-10-20T10:31:04.210

12A more elementary, and I would bet more common, mistake is to believe that exp(A+B)=exp(A) exp(B) with no hypotheses on A and B. – David E Speyer – 2010-09-27T13:40:04.833


The gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=0,-1,-2\dots$.

In fact, there is a whole bunch of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\qquad g(1)=\gamma+2l\pi i,\qquad k,l\in\mathbb Z, $$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.

Andrey Rekalo

Posted 2010-05-04T21:02:58.510

Reputation: 18 511

I think I remember reading it is the only one that is convex over $(0,+\infty)$ or some similar condition. – Arnaud Chéritat – 2015-10-18T12:45:50.477

3It is the only one that is log convex (ie, $\log(\Gamma(x))$ is a convex function). – Rellek – 2017-03-14T19:07:33.133


Here's one that bugged me from point set topology: "A subnet of a sequence is a subsequence".

See here for the definitions. Using this one gives a great proof that compactness implies sequential compactness in any topological space:

Let $X$ be a compact space. Let $(x_n)$ be a sequence. Since a sequence is a net and it's a basic theorem of point set topology that in a compact topological space, every net has a convergent subnet (proof in the above link), there is a convergent subnet of the sequence $(x_n)$. Using the above belief, the sequence $(x_n)$ has a convergent subsequence and hence $X$ is sequentially compact.

For a counterexample to this "theorem", consider the compact space $X= \lbrace 0,1 \rbrace ^{[0,1]}$ with $f_n(x)$ the $n$th binary digit of $x$.

Anthony Quas

Posted 2010-05-04T21:02:58.510

Reputation: 14 900

I've just added the missing html to fix the link. – gowers – 2011-07-03T21:38:38.607


The product of two symmetric matrices is symmetric!


Posted 2010-05-04T21:02:58.510

Reputation: 1 521

1Surprising! I explored a little and discovered that if the matrices are instead radially symmetric then the product is as well. That is, aij = a(n - i)(n - j). – Whosyourjay – 2016-03-06T07:29:34.917


The ring $\mathbb{C}[x]$ has countable dimension over $\mathbb{C}$; therefore its field of fractions $\mathbb{C}(x)$ also has countable dimension over $\mathbb{C}$.

Todd Trimble

Posted 2010-05-04T21:02:58.510

Reputation: 41 195

is this not true? – Turbo – 2014-11-29T22:27:11.803

6The uncountably many elements $1/(x-a)$ for all $a \in \mathbb C$ are linearly independent. – darij grinberg – 2014-11-29T22:47:59.443

By the way, this fact is the basis for a beautiful proof of Hilbert's Nullstellensatz. – ACL – 2016-04-21T06:40:21.210

@ACL I learned this not very long ago myself, from here: http://mathoverflow.net/a/15232/2926

– Todd Trimble – 2016-04-21T07:45:40.883

Wow, good one. For by me now it is very viscerally known that "passage to the field of fractions does not change the size." This is true in the sense of cardinality, but.... – Pete L. Clark – 2017-09-06T22:16:03.940

This fact is used to prove Schur's lemma for l-group which is countable at infinity. (In the language of Bernstein-Zelevinsky) – Qing Zhang – 2017-09-16T12:37:27.790

However, this works for finite ring extensions (see here).

– Watson – 2017-10-17T17:43:23.233


Here's one that doesn't seem to be listed here: That ZFC isn't vulnerable to Russell's paradox because it has an axiom which implies that sets cannot be elements of themselves.

This one should be obviously false given even a moment's thought -- you cannot get rid of a paradox by adding axioms! But I've seen it repeated over and over.

The correct statement, of course, is that ZFC isn't vulnerable to Russell's paradox because it doesn't have the axiom of unrestricted comprehension; and that while ZFC does have an axiom which implies that all sets are well-founded, this is irrelevant to Russell's paradox. (It just means that the set of all sets which are not elements of themselves, if it existed, would happen to also be the set of all sets.)

Sometimes this is stated in an ambiguous way that allows for the charitable reading that modern set theory prevents Russell's paradox by preventing one from talking about a set being an element of itself, which is how Russell and Whitehead fixed the problem. But Principia Mathematica isn't exactly "modern set theory"; and I've seen this stated the blatantly wrong way often enough that I'm not too inclined to be charitable about it.

Harry Altman

Posted 2010-05-04T21:02:58.510

Reputation: 1 352

3Why can't you get rid of a paradox by adding axioms? Axioms can certainly constrain as well as construct. – Steven Stadnicki – 2016-12-22T05:15:31.290

10Because if you establish inconsistency with a proof which uses only axioms $A_1, \ldots, A_n$ in the old system, and add an axiom $A$, then you still have the old proof which used only the axioms $A_1, \ldots, A_n$. Adding axioms increases the potential for proving inconsistency; it can never decrease it. – Todd Trimble – 2017-07-13T20:10:05.700


I and several of my friends made our living off exploiting the fact that a space doesn't really have just one universal cover.

It has one at each basepoint. If the space is decent and connected, then these are all isomorphic, but the isomorphism requires a choice of a path connecting the points. You can clearly see the point if you make a bundle of universal covers over a moebius strip.

shmuel weinberger

Posted 2010-05-04T21:02:58.510

Reputation: 1

4@DavidRoberts : What he does for a living is that he makes a bundle off universal covers. $\qquad$ – Michael Hardy – 2016-03-10T05:13:36.343

1@MichaelHardy how droll :-) – David Roberts – 2016-03-10T06:11:28.837

21You make a living off it?? Awesome. What do you do? – David Roberts – 2010-08-31T21:43:39.390

3Related to this, here is a cool trick question which I heard from Allen Knutson: Consider the Hopf fibration, which is an $S^1$ bundle over $S^2$. Take the universal cover of each fiber, to get an $\mathbb{R}$ bundle over $S^2$. Is this line bundle nontrivial? – David E Speyer – 2010-09-27T13:38:03.290


A projection of a measurable set is measurable. Not only students believe this. I was asked once (the quote is not precise): "Why do you need this assumption of a measurability of projection? It follows from ..."

A polynomial which takes integer values in all integer points has integer coefficients.

Another one seems to be more specific, I just recalled it reading this example. A sub-$\sigma$-algebra of a countably generated $\sigma$-algebra is countably generated.


Posted 2010-05-04T21:02:58.510

Reputation: 1 107

A counterexample to the polynomial thing is $\frac12x^2+\frac12x$. – Akiva Weinberger – 2015-08-31T22:27:07.073


Another false belief which I have been asked thrice so far in person is $$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$ even if $x$ is in degrees. I was asked by a student a year and half back when I was a TA and by couple of friends in the past 6 months.


Posted 2010-05-04T21:02:58.510


4maybe not one of the best answers here, but why the down votes? – Yaakov Baruch – 2011-02-23T15:08:19.860

3@downvoters: Kindly provide a reason for the down votes. – None – 2011-02-23T15:54:55.800

1+1 from me. I will try to keep this potential source of confusion in mind until I teach on this the next time. – None – 2011-02-24T17:47:43.013

1I can't believe one actually can believe this... – darij grinberg – 2011-02-24T22:54:33.297


+1. The limit when $x$ is in degrees is an exercise in many calculus textbooks (or equivalently, the derivative of $\sin (x degrees)$. Yet, it seems people are slow to pick up on it. Your point was made by Deane Yang in this answer: http://mathoverflow.net/questions/40082/why-do-we-teach-calculus-students-the-derivative-as-a-limit/40136#40136

(and no one found anything wrong with it then...)

– Thierry Zell – 2011-02-27T14:45:49.710

1Probably, it received down-votes because the question says, "The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements...)." Certainly, I'm inclined to down-vote on that basis. – JBL – 2011-03-06T04:40:02.463

1@Sivaram: I think it received down votes because noone likes degrees. 360 is quite an arbitrary choice. – Eric Naslund – 2011-03-06T06:54:02.707

@JBL: Probably... I however feel that it is equally elementary to think that $\sin(z)$ is bounded in the complex domain. – None – 2011-03-06T08:04:32.093

1Sivaram, I guess you are complaining that some other answer wasn't downvoted (but I'm not sure because I'm certainly not going to go back and look through all past answers to check; but anyway how would you know it hadn't been down-voted and then later up-voted, as this one)? But the second example is obviously less elementary -- the first one is taught to every high school student or college freshman who takes calculus. The phrase "sin(x) is bounded in the complex domain" is incomprehensible to anyone not in a first complex analysis course. – JBL – 2011-03-06T15:08:36.047

2To my knowledge, there is only one sine function, and it is a map from $\mathbb{R}$ to $\mathbb{R}$ (or from $\mathbb{C}$ to $\mathbb{C}$, if you insist). What does it mean for a real number to be "in degrees"? And if you redefine the sine function as a map from $A$ to $\mathbb{R}$ where $A$ is the space of "angles" (whatever this means), then $\frac{\sin\,(x)}{x}$ is meaningless. – Laurent Moret-Bailly – 2011-03-06T16:13:43.630

2@JBL, what Sivaram says is taken directly from the question, an example of what is asked for. Granted, this is slightly more advanced. Yet, the second example given 'open dense sets in R' is (in certain uni-curricula) something that comes up earlier than sin (at the level of rigor needed to talk about limits). @Laurent Moret-Bailly, yes and no: define sind(x)= sin(pi x /180), to ask what the limit of sind(x)/x is is not meaningless. And, on varios calculators pressing 'sin' gives this 'sind' (or at least they have that option). – None – 2011-03-11T16:01:34.673

@unknown, students in the US learn trigonometric functions some time in high school and learn limits in a first calculus course, typically in the last year of high school or first year of college. The words "open" and "dense" are not defined in any of these courses, except maybe that some students know what an open interval is. – JBL – 2011-03-11T16:26:14.573

4@JBL: Well, there are also some universities outside the US ;) This is not standard, yet not unusual though becoming rarer, in certain parts of Europe: In HS one learns about trig. func. in a geom. way; about diff./int. without a formal notion of limit, mainly rat. funct; in any case that limit wouldn't show up explictly. (Maybe 'invisibly' if derivative of trig. functions are mentioned.) Then, at univ. at the very start you take (real) analysis: constr. of the reals, basic top. notions(!), continuity,...,series of functions as application powerseries, and as appl exp and trig. func. – None – 2011-03-11T17:47:38.353

4@Laurent Moret-Bailly:

The definition of $\frac{\sin x}x$ in degrees is the number you get when you type it into your calculator while forgetting to push deg/rad/grad first. – user11235 – 2011-04-10T20:41:38.853


A subgroup of a finitely generated group is again finitely generated.


Posted 2010-05-04T21:02:58.510

Reputation: 138

1True for abelian groups, though. – Mark – 2011-03-03T21:40:15.903

1Also true for finite index subgroups, of finitely generated groups. – Michalis – 2011-03-04T21:18:41.817


I've already mentioned this in my answer http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23640#23640, but nevermind. ;)

– Martin Brandenburg – 2011-04-12T08:44:34.620

I had this false belief until the 2nd year of grad school. – Michael – 2013-12-03T01:21:56.617


My example is $G_{1}$ isomorphic to $G_{2}$'s subgroup and $G_{2}$ isomorphic to $G_{1}$'s subgroup implies $G_{1}$ and $G_{2}$ are isomorphic...


Posted 2010-05-04T21:02:58.510

Reputation: 465


See also http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold .

– Qiaochu Yuan – 2010-05-06T00:53:29.553

16I'd think of that not as a belief that people are likely to have so much as a statement that looks moderately plausible and doesn't have an obvious counterexample. In other words, it's not something that people unthinkingly assume, because they don't tend to think about it at all. But perhaps I'm wrong about that and it is routinely used as a lemma by inexperienced algebraists. – gowers – 2010-05-06T11:43:50.193


This is (I think) a fairly common misconception about maths that arises in connection with quantum mechanics. Given a Hermitian operator A acting on a finite dimensional Hilbert space H, the eigenvectors of A span H. It's easy to think that the infinite dimensional case is "basically the same", or that any "nice" operator that physicists might want to consider has a spanning eigenspace. However, neither the position nor the momentum operator acting on $L^2(\mathbb{R})$ have any eigenvectors at all, and these are certainly important physical operators! Based on an admittedly fairly small sample size, it seems that it's not uncommon to simultaneously believe that Heisenberg's uncertainty relation holds and that the position and momentum operators possess eigenvectors.

Phil Ellison

Posted 2010-05-04T21:02:58.510

Reputation: 148

1Yeah, for some reason many physicists are taught exactly no functional analysis... In fact, I know of no "quantum mechanics for physicists" books which use much more than a beginning undergrad level of analysis. Though admittedly these details are not so important for doing simple calculations, though they can be important in doing more sophisticated calculations, or understanding, e.g., why field theory works the way it does... – jeremy – 2010-06-01T23:33:12.630

7Reciprocally, many mathematicians are taught no quantum mecha... make it, no physics at all! This is shocking, since the biggest impetus to the development of PDEs and functional analysis was given by what? You guessed it, physics. – Victor Protsak – 2010-06-10T06:56:20.283


Perhaps the most prevalent false belief in math, starting with calculus class, is that the general antiderivative of f(x) = 1/x is F(x) = ln|x| + C. This can be found in innumerable calculus textbooks and is ubiquitous on the Web.

Daniel Asimov

Posted 2010-05-04T21:02:58.510

Reputation: 856

In a related vein, before discussing improper integrals, some books ask students to evaluate the riemann integral of functions like 1/sqrt(x), on [0,1], without noting that the function is unbounded hence not riemann integrable. the fact that the antiderivative exists and is finite on [0,1] masks this problem. – roy smith – 2011-05-09T02:53:03.737

5Well, the false belief is correct under the (frequently unspoken) condition that we only speak of antiderivatives over intervals on which the function we're antidifferentiating is "well-behaved" (and I'm not 100% sure what the right technical condition there is; "continuous"?). – JBL – 2010-06-12T00:57:01.390

6Really? What about the function F(x) given by

   ln(x)  + C_1,  x &gt; 0

F(x) = ln(-x) + C_2, x < 0

for arbitrary reals C_1, C_2 ?

(The appropriate technical condition is that an antiderivative be differentiable on the same domain as the function it's the antiderivative of is defined on.) – Daniel Asimov – 2010-06-12T04:25:16.277

4In case that wasn't clear: F(x) = ln(x) + C_1 for x > 0, and F(x) = ln(-x) + C_2 for x < 0, where C_1 and C_2 are arbitrary real constants. – Daniel Asimov – 2010-06-12T04:29:34.400

8That function is not "nice" on any interval containing 0; on any interval not containing 0, it is of the form you are complaining about. This is exactly my point -- the word "interval" is important to what I wrote! – JBL – 2010-06-12T19:33:36.440

This false belief is self-consistent, hence irrefutable, as far as antidifferentiation is concerned, because one can always choose the (true) different local constants of integration to match the (false but notationally implied) single global constant. It is difficult or impossible to find an elementary function with singularities, that can be formally anti-differentiated in two different ways (incorrectly using a single "global" integration constant at each step), where subtracting one antiderivative from the other produces a locally-but-not-globally constant function. – T.. – 2010-06-12T20:45:43.590

3Sorry, I'm surely communicating less clearly than would be ideal. Starting over from scratch: suppose we have some differential equation, with unknown function $y(x)$. We say that for some function $f(x)$, $y = f(x)$ is a solution of the differential equation if there exists an interval $I$ on which $f(x)$ has all the requisite derivatives and the equation is satisfied on this interval. Antidifferentiation is the particular case $y' = g(x)$; implicit in the statement "$F(x)$ is an antiderivative of $f(x)$" is the condition "on some interval for which $F(x)$ is differentiable." – JBL – 2010-06-12T22:02:00.473

2Now, if your point is that most students (and perhaps many instructors) in calculus classes often don't realize that this condition is necessary when they write $\int \frac{1}{x} \, dx = \ln |x| + C$, I agree with you; and if your point is that it would be less misleading to say something like, "For $x > 0$, the antiderivative of $\frac{1}{x}$ is $\ln x + C$, while for $x < 0$ the antiderivative of $\frac{1}{x}$ is $\ln (-x) + C$," I also agree with you; and if your point is that many people might not realize that there are functions (cont'd) – JBL – 2010-06-12T22:05:42.793

3$\mathbb{R}^\times \to \mathbb{R}$ other than $\ln |x| + c$ with derivative $\frac{1}{x}$, I also agree with you; I just happen to think that the actual statement you wrote down is not incorrect but rather has an unwritten assumption built into the word "antiderivative," namely that such a thing is only defined for an interval on which the supposed antiderivative is differentiable.

I hope this is clearer (and also correct!). – JBL – 2010-06-12T22:13:17.203


Jonas Meyer

Posted 2010-05-04T21:02:58.510

Reputation: 5 808

6This false belief is perhaps caused by the fact that continuity does imply sequential continuity, and sequential adherent points are adherent points. – Terry Tao – 2010-09-27T05:53:54.970


By definition, an asymptote is a line that a curve keeps getting closer to but never touches. The teaching of this false belief at an elementary level is standard and nearly universal. Everybody "knows" that it is true. A tee-shirt has a clever joke about it. In the course of describing the function $f(x) = \dfrac{5x}{36 + x^2}$, I mentioned about an hour ago before a class of about 10 students that its value at 0 is 0 and that it has a horizontal asymptote at 0. One of them accused me of contradicting myself. What of $y = \dfrac{\sin x}{x}$? And even with simple rational functions there are exceptions, although there the curve can touch or cross the asymptote only finitely many times. And $3 - \dfrac{1}{x}$ gets closer to 5 as $x$ grows, and never reaches 5, so by the widespread false belief there would be a horizontal asymptote at 5.

Michael Hardy

Posted 2010-05-04T21:02:58.510

Reputation: 5 412

+1: I actually had this exact misconception several years ago!! I cannot remember the particular example, but I know I went to my teacher and told him the function could not have an asymptote since it crossed the line. – Eric Naslund – 2011-03-06T06:58:10.533

6For this to be a false definition, it would have to be a definition in the first place. And this means you have to define a "curve" first, and then define "get closer" and "touch". – Laurent Moret-Bailly – 2011-03-06T16:01:06.337

10@Laurent: It's hard to imagine a comment more irrelevant to what happens in classrooms than yours. – Michael Hardy – 2011-03-07T04:38:27.687

7It happens to be the literal meaning of the word asymptote "not together falling". You could say that it is a bad choice of name, but for hyperbolas it worked just fine and then it was mercilessly generalized. – user11235 – 2011-04-08T14:35:14.240


Regard a reasonably nice surface in $\mathbb R^3$ that can locally be expressed by each of the functions $x(y,z)$, $y(x,z)$ and $z(x,y)$, then obviously

$\frac {dy} {dx} \cdot \frac {dz} {dy} \cdot \frac {dx} {dz} = 1$

(provided everything exists and is evaluated at the same point).

After all, this kind of reasoning works in $\mathbb R^2$ when calculating the derivative of the inverse function, it works for the chain rule and it works for separation of variables.

Note that this product is in fact $-1$ which can either be seen by just thinking about what happens to the equation $ax+by+cz=d$ of a plane / tangent plane or by looking at the expression coming out of the implicit function theorem.

I recall someone claiming that this example proves that $dx$ should be regarded as linear function rather than infinitesimal, but I cannot reconstruct the argument at the moment as this discussion was 15 years ago.

In particular, it is true under appropriate conditions in $\mathbb R^4$ that $\frac {\partial y} {\partial x} \cdot \frac {\partial z} {\partial y} \cdot \frac {\partial w} {\partial z} \cdot \frac {\partial x} {\partial w} = 1$


Posted 2010-05-04T21:02:58.510

Reputation: 453

10This is an example of the principle that naïve reasoning with Leibniz notation works fine for total derivatives but not for partial derivatives. This is one reason why I would always write the left-hand side as

$\frac{\partial{y}}{\partial{x}} \cdot \frac{\partial{z}}{\partial{y}} \cdot \frac{\partial{x}}{\partial{z}}$

if not

$\left(\frac{\partial{y}}{\partial{x}}\right)_z \cdot \left(\frac{\partial{z}}{\partial{y}}\right)_x \cdot \left(\frac{\partial{x}}{\partial{z}}\right)_y$

(notation that I learnt from statistical physics, where the independent variables are otherwise not clear). – Toby Bartels – 2011-04-07T12:56:12.480

But this notation does not help one to understand that the above expression is actually $-1$. – user11235 – 2011-04-10T11:05:46.587

3Can you help us understand it? Or is there no better way than computation? – darij grinberg – 2011-04-10T18:27:28.880

3@TobyBartels, I remember an analyst colleague talking about a problem in a paper of his that he resolved by noticing (if I remember the particular example correctly) that $\partial/\partial r$ means something different in cylindrical and spherical co\"ordinates. – LSpice – 2013-12-12T23:17:32.347


Just today I came across a mathematician who was under the impression that $\aleph_1$ is defined to be $2^{\aleph_0}$, and therefore that the continuum hypothesis says there is no cardinal between $\aleph_0$ and $\aleph_1$.

In fact, Cantor proved there are no cardinals between $\aleph_0$ and $\aleph_1$. The continuum hypothesis says there are no cardinals between $\aleph_0$ and $2^{\aleph_0}$.

$2^{\aleph_0}$ is the cardinality of the set of all functions from a set of size $\aleph_0$ into a set of size $2$. Equivalently, it is the cardinality of the set of all subsets of a set of size $\aleph_0$, and that is also the cardinality of the set of all real numbers.

$\aleph_1$, on the other hand, is the cardinality of the set of all countable ordinals. (And $\aleph_2$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_1$, and so on, and $\aleph_\omega$ is the next cardinal of well-ordered sets after all $\aleph_n$ for $n$ a finite ordinal, and $\aleph_{\omega+1}$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_\omega$, etc. These definitions go back to Cantor.

Michael Hardy

Posted 2010-05-04T21:02:58.510

Reputation: 5 412

Do you think this is common? – None – 2011-10-05T16:40:33.887

3I retract my above question to my suprise it indeed seems to be common. Yet, this answer is a dublicate see an answer of April 16. – None – 2011-10-06T00:50:05.097


This example already appears on this very page. http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/61975#61975

– Asaf Karagila – 2011-10-06T12:41:21.827

1One of the deficiencies of mathoverflow's software is that there is no easy way to search through the answers already posted. Even knowing that the date was April 16th doesn't help. – Michael Hardy – 2011-10-07T20:26:28.473

@quid: How did you reach your conclusion that this is commonplace, after your initial doubts? – Michael Hardy – 2011-10-07T20:27:25.323

After all two people gave the answer ;) [And, needless to say, I was unaware of this when writing my first comment.] Also the other answer says something that it happens to him somehow frequently; while you wrote it like it happened the first time to you. Finally, for its relatively young age that other answer has many upvotes, so people seem to agree it is common, and here voting should be reliable. Personally, I though it would not be common: one either never heard of aleph's or knows what they are (so no room for false belief); but perhaps elsewhere the term. is more common. – None – 2011-10-07T20:50:45.920

Actually I don't think it was the first time, but it was the first time since this question was posted on mathoverflow. – Michael Hardy – 2011-10-07T22:20:29.137

2@Michael Hardy: You can sort the answers by date by clicking on the "Newest" or "Oldest" tabs instead of the "Votes" tab. – Douglas Zare – 2011-10-19T23:03:30.293

Incidentally, Isaac Asimov's writings on mathematics are usually correct and good for young adults, but he did not get all of the subtleties about the continuum hypothesis correct. See page 140 http://www.scribd.com/doc/66612470/Realm-of-Numbers-Issac-Asimov. I don't recall if he made this precise error explicitly but I had this misconception after reading Asimov On Numbers.

– Douglas Zare – 2011-10-19T23:15:22.053

1I suspect much confusion stems originally from George Gamow's book "One Two Three ... Infinity" — which got a number of things wrong about the continuum and the continuum hypothesis, which it implied was settled. None of the reprintings of this book fixed the error — to this day. – Daniel Asimov – 2017-05-29T16:22:24.790


  • The category of commutative C*-algebras is equivalent to the opposite category of locally compact Hausdorff spaces.

It's actually not quite that simple! There is some discussion on math.SE.

Tobias Fritz

Posted 2010-05-04T21:02:58.510

Reputation: 1 396

Very interesting point. – Ali Taghavi – 2014-11-12T12:42:34.333

3Whoops! Thanks for posting this and particularly the link, which shows that the nLab got this one wrong as well. – Todd Trimble – 2014-11-12T13:03:32.557


"Some real numbers are not definable, by Cantor's diagonal argument."

There are subtleties involved in formalizing the statment "some real numbers are not definable", as explained in Joel's answer to this question. The statement can be seen to hold in some models and fail in other models. However, the claim that the statement follows from Cantor's diagonal argument is clearly false, yet seems to be fairly common.

The false reasoning typically proceeds in three steps:

  1. There are only countably many definitions of real numbers: $\varphi_0(x),\varphi_1(x),\ldots$ (this part is ok.)

  2. Consider the countably many real numbers so defined: $x_0,x_1,\ldots$ (this part is problematic for subtle reasons.)

  3. Use Cantor's diagonal argument to obtain a real number $y$ that is not in the sequence from step 2, and is therefore not definable.

For the moment, let us assume that step 2 succeeds in the way that one might naively think it would. Then we have defined a sequence $x_0,x_1,\ldots$ containing all definable real numbers. Therefore Cantor's diagonal argument in step 3 defines, from this sequence, a real number $y$ that is not in the sequence. So $y$ is both definable and not definable, and we obtain a contradiction outright! Clearly, something is wrong (and it turns out to be in step 2.)

Trevor Wilson

Posted 2010-05-04T21:02:58.510

Reputation: 2 863

Is it because ${x_0,x_1,\dots}$ isn't necessarily a set? – Akiva Weinberger – 2015-09-01T00:17:57.053

@columbus8myhw It's related to that, although it's possible for ${x_0, x_1, \ldots}$ to be a set for unrelated reasons. (For example, given a pointwise definable model of set theory, we can consider ${x_0, x_1, \ldots}$ from the outside, and see that it happens to equal a set in the model, namely $\mathbb{R}$ itself.) But in this case $(x_0,x_1,\ldots)$ will fail to be a sequence (of the model) and the argument will still fail (in the model). – Trevor Wilson – 2015-09-01T18:30:40.547

That being said, understanding why ${x_0,x_1,\ldots}$ isn't necessarily a set is similar to understanding why the argument isn't valid. – Trevor Wilson – 2015-09-01T18:31:34.787


To my knowledge, noone has proven that the scheme of pairs of matrices (A,B) satisfying the equations AB=BA is reduced. But whenever I mention this to people someone says "Surely that's known to be reduced!"

(Similar-sounding problem: consider matrices M with $M^2=0$. They must be nilpotent, hence have all eigenvalues zero, hence $Tr(M)=0$. But that linear equation can't be derived from the original homogeneous quadratic equations. Hence this scheme is not reduced.)

Allen Knutson

Posted 2010-05-04T21:02:58.510

Reputation: 20 250

I remember Etingof thinking it was open as recently as 2006 in his lectures on Calogero-Moser systems. Also, it's fairly easy to check that in small cases the commutator scheme is reduced, though getting it in general seems to be beyond current technology. – Charles Siegel – 2010-06-06T21:13:21.107

2Sadly, people rely on technology so much nowadays that it gets increasingly unlikely that it will $\textit{ever}$ be proved. – Victor Protsak – 2010-06-10T06:40:27.647


A common false belief is that all Gödel sentences are true because they say of themselves they are unprovable. See Peter Milne's "On Goedel Sentences and What They Say", Philosophia Mathematica (III) 15 (2007), 193–226. doi:10.1093/philmat/nkm015.

Marc Alcobé García

Posted 2010-05-04T21:02:58.510

Reputation: 627


In his answer above, Martin Brandenburg cited the false belief that every short exact sequence of the form

$$0\rightarrow A\rightarrow A\oplus B\rightarrow B\rightarrow 0$$

must split.

I expect that a far more widespread false belief is that such a sequence can fail to split, when A, B and C are finitely generated modules over a commutative noetherian ring.

(Sketch of relevant proof: We need to show that the identity map in $Hom(A,A)$ lifts to $Hom(A\oplus B,A)$. Thus we need to show exactness on the right of the sequence

$$0\rightarrow Hom(B,A)\rightarrow Hom(A\oplus B,A)\rightarrow Hom(A,A)\rightarrow 0$$

For this, it suffices to localize and then complete at an arbitrary prime $P$. But completion at $P$ is a limit of tensorings with $R/P^n$, so to check exactness we can replace the right-hand $A$ in each Hom-group with $A/P^nA$. Now we are reduced to looking at modules of finite length, and the sequence is forced to be exact because the lengths of the left and right terms add up to the length in the middle. This is due, I think, to Miyata.)

Steven Landsburg

Posted 2010-05-04T21:02:58.510

Reputation: 15 393

1Very interesting! – Martin Brandenburg – 2011-04-12T08:48:32.170


It always confused me as an undergraduate that $\mathbb{Q}\subset\mathbb{R}$ has an open neighborhood $N_\epsilon \supset\mathbb{Q}$ of arbitrarily small measure $\epsilon$, because $\mathbb{Q}$ is dense in $\mathbb{R}$.


Posted 2010-05-04T21:02:58.510

Reputation: 288

(This answer is very similar to Owen Sizemore's answer that is currently on the first page.) – R. van Dobben de Bruyn – 2017-07-24T14:41:14.727


Here's a little factoid: (The Mean-value theorem for functions taking values in $\mathbb{R} ^n$.) If $\alpha : [a,b]\rightarrow \mathbb{R}^n$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in (a,b)$ such that $\frac{\alpha (b)-\alpha (a)}{b-a}=\alpha '(c)$

A counterexample is the helix $(\cos (t),\sin (t), t)$ with $a=0$, $b=2\pi$.

Another common misunderstanding (although not mathematical) is about the meaning of the word factoid. In fact, the common mistaken definition of the word factoid is factoidal.

Robin Tucker-Drob

Posted 2010-05-04T21:02:58.510

Reputation: 429

Related to the M-V Thm is the following fact. If $f:I=(a,b)\rightarrow{\mathbb R}$ is differentiable (not necessarily ${\mathcal C}^1$), then $f'(I)$ is connected (i.e. is an interval). This is false when $f:I=(a,b)\rightarrow{\mathbb R}^n$ is differentiable, and $n\ge2$. – Denis Serre – 2010-10-20T10:48:15.423

10On the other hand, perhaps the most useful corollary of the mean value theorem is the "mean value inequality": that $|\alpha(b) - \alpha(a)| \le (b-a) \sup_{t \in [a,b]} |\alpha'(t)|$. If you look carefully, most applications of the MVT in calculus are really using this "MVI". The MVI remains true for absolutely continuous functions taking values in any Banach space, and so is probably the right generalization to keep in mind. – Nate Eldredge – 2010-05-06T14:37:12.743

1According to at least one dictionary, there are two different definitions of factoid: (1) an insignificant or trivial fact, and (2) something fictitious or unsubstantiated that is presented as fact, devised especially to gain publicity and accepted because of constant repetition. I am not convinced that the multi-d mean value “theorem” fits either definition. – Harald Hanche-Olsen – 2010-05-08T19:09:03.027


In the past I have found myself making this mistake (probably fueled by the fact that you can indeed extend bounded linear operators), and I think it is common in students with a not-deep-enough topology background:

"Let $T$ be a compact topological space, and $X\subset T$ a dense subset. Take $f:X\to\mathbb{C}$ continuous and bounded. Then $f$ can be extended by continuity to all of $T$ ".

The classical counterexample is $T=[0,1]$, $X=(0,1]$, $f(t)=\sin\frac1t$ . It helps to understand how unimaginable the Stone-Cech compactification is.

Martin Argerami

Posted 2010-05-04T21:02:58.510

Reputation: 1 846

14How about this one: $T=[-1,1]$, $X=T-{0}$, $f(x)=$ sign of $x$. – Laurent Moret-Bailly – 2010-10-19T07:23:52.720

2Nice! That's certainly a much simpler example. – Martin Argerami – 2010-10-19T10:45:05.110

3Indeed; the key property is uniform continuity. – Nate Eldredge – 2010-10-14T14:37:42.860


That Darboux functions are continuous is certainly a widely held belief among students, at least in France where it is induced by the way continuity is taught in high school.

I remember having gone through all the five "stages of grief" when shaken from this false belief with the $\sin(1/x)$ example : denial, anger ( "then the definition of continuity must be wrong ! Let's change it !), bargaining ("Ok, but a Darboux function must surely be continuous except at exceptional points. Let's prove that..."), depression (when shown a nowhere continuous Darboux function), acceptance ("Hey guys, you really think the intermediate value theorem has a converse? C'mon, you're smarter than that...")

Fabien Besnard

Posted 2010-05-04T21:02:58.510

Reputation: 160


The distinction between convergence and uniform convergence. It even got Cauchy in its time.

Anna Taurogenireva

Posted 2010-05-04T21:02:58.510

Reputation: 368


  • Many students have the false belief that if a topological space is totally disconnected, then it must be discrete (related to examples already given). The rationals are a simple counter-example of course.

  • It is common to imagine rotation in an n-dimensional space, as a rotation through an "axis". this is of course true only in 3D, In higher dimensions there is no "axis".

  • In calculus, I had some troubles with the following wrong idea. A curve in a plane parametrized by a smooth function is "smooth" in the intuitive sense (having no corners). the curve that is defined by $(t^2,t^2)$ for $t\ge0$ and $(-t^2,t^2)$ for $t<0$ is the graph of the absolute value function with a "corner" at the origin, though the coordinate functions are smooth. the "non-regularity" of the parametrization resolves the conflict.

  • When first encountering the concept of a spectrum of a ring, the belief that a continuous function between the spectra of two rings must come from a ring homomorphism between the rings.


Posted 2010-05-04T21:02:58.510

Reputation: 852

What do you mean by "In higher dimensions there is no "axis" but a n-2 dimensional subspace instead" ? Whenever n is even, there are rotations without real eigenvectors. – Johannes Hahn – 2011-04-14T13:12:54.127

2Unfortunately, "smooth" is a word which means whatever its utterer does not want to specify. Differentiable, C^infty, continuous, everything is mixed. – darij grinberg – 2011-04-14T15:12:48.150

@Zsban: thanks for noticing. I edited the post. now I believe it is correct (though less impressive perhaps...). – KotelKanim – 2011-05-03T15:25:50.690

2+1 for the discrete $\neq$ totally disconnected example. – Jim Conant – 2011-05-04T15:12:41.257

It's still possible to have an example where the parametric definition is a smooth (as in $ C^{\infty} $) function of time but you get a corner, right? – Zsbán Ambrus – 2011-05-09T14:46:31.493

@Zsban: yes of course, you will just have to use a bit more complicated formula, but it can be done by the "table function" for example. the point is that you can slow down smoothly until you actually stop at the corner and then you start to accelerate smoothly towards a different direction. – KotelKanim – 2011-05-10T19:10:02.950

2Discrete $\ne$ totally disconnected is a good one that I thought of today and just had to check to see if it was posted already. It adds to the confusion that every finite subset of a totally disconnected space must have the discrete topology, and that in most topological spaces encountered "in nature," the connected components are open sets. – Timothy Chow – 2011-10-20T14:30:41.957


In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\ $ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).

Igor Pak

Posted 2010-05-04T21:02:58.510

Reputation: 11 543

1Oh, I previously thought that eight regular tetrahedra with side length 1 fit in one with side length 2. That might be a more common false belief. – Junyan Xu – 2012-05-05T07:45:29.587


A stunning, ignorance-based false belief I have witnessed while observing a class of a math education colleague is that there is no general formula for the n-th Fibonacci number. I wonder if this false belief comes from conflating the (difficult) lack of formulas for prime numbers with something that is just over the horizon of someone whose interests never stretch beyond high-school math.

Behind a number of the elementary false beliefs listed here there is a widespread tendency among people to give up too easily (maybe when having to read at least to page 2 in a book), or to nourish an ego that allows to conclude that something is impossible if they cannot do it themselves.

Vania Mascioni

Posted 2010-05-04T21:02:58.510

Reputation: 91

2Perhaps the people who believe this are using the meta-reasoning that the sequence would not be interesting as an example of recursion if it could be solved exactly. Since it is a popular example of recursion, then... – Ryan Reich – 2011-10-05T17:01:55.040

1When, as an undergrad, I couldn't solve a problem given to me by the advisor, and asserted that it's "unsolvable", the advisor replied that "solvability of a problem is a function of two arguments: the problem and the solver." – Michael – 2013-12-03T01:14:39.963

@Michael (The constant function is still a function. :P ) – Akiva Weinberger – 2015-09-01T00:03:31.730

2I hope at least your colleague had it right! There is another one along these lines: there is no formula for the sequence $1,0,1,0,1,0,... .$ Your second paragraph is right on target, but I also think that the specific beliefs you and I mentioned have a lot to do with a very limited understanding of what is a "formula". – Victor Protsak – 2010-06-10T07:02:45.270


There are cases that people know that a certain naive mathematical thought is incorrect but largely overestimate the amount by which it is incorrect. I remember hearing on the radio somebody explaining: "We make five experiments where the probability for success in every experiment is 10%. Now, a naive person will think that the probability that at least one of the experiment succeed is five times ten, 50%. But this is incorrect! the probability for success is not much larger than the 10% we started with."

Of course, the truth is much closer to 50% than to 10%.

(Let me also mention that there are various common false beliefs about mathematical terms: NP stands for "not polynomial" [in fact it stands for "Nondeterministic Polynomial" time]; the word "Killing" in Killing form is an adjective [in fact it is based on the name of the mathematician "Wilhelm Killing"] etc.)

Gil Kalai

Posted 2010-05-04T21:02:58.510

Reputation: 12 545

Assuming the experiments are independent events, I get that the probability for success in at least one of them is $40.9 \%$. I would call that "much larger than the $10\%$ we started with". Am I making a mistake? – Pete L. Clark – 2010-05-05T11:34:56.030

Right! but the radio commentatior was so happy that he know that it is not 50% that he falsely assumed "it is not much larger than 10%". (I added a sentence to clarify.) – Gil Kalai – 2010-05-05T11:43:55.283

OK, I get it now. I was misinterpreting where the commentator ended and your commentary began. The quotation remarks you added are very helpful. – Pete L. Clark – 2010-05-05T11:53:29.180

10And the Killing field has nothing to do with Pol Pot. – Nate Eldredge – 2010-05-05T14:40:41.690

I certainly held the belief that the Killing form is named after Wilhelm Killing, but my belief is based on the incontrovertible evidence that this statement appears in Wikipedia:


Is Wikipedia for sure wrong?

– Paul Siegel – 2010-05-05T16:37:28.920

@Paul Siegel and Gil Kalai: I rewrote the last comment in the post because it was causing confusion with a mixture of positive and negative statements. – Willie Wong – 2010-05-05T17:13:54.250

3Unfortunately I often slip up in class and say that the Killing vector field $T$ kills the metric term (well, I use the verb kills when a differential operator hits something and makes it zero, because, you know, bad terms are always "the enemy"). I'm not sure how much damage I did to the students' impressions... – Willie Wong – 2010-05-05T17:19:39.663

3"Kills" is one of those terms I hear mathematicians use surprisingly often. The other one is "this guy." I never really understood the prevalence of either. – Qiaochu Yuan – 2010-05-06T07:38:20.733

22"Guy" is a pretty standard English colloquialism for "person"; combine this with humans' tendency to anthropomorphize and this usage is understandable. (Though we shouldn't anthropomorphize mathematical objects, because they hate that.) – Nate Eldredge – 2010-05-06T14:51:01.073

2"This guy" is one of the reasons I stopped taking notes in math classes. Third year in college, abstract algebra, a typical sentence in the lecture was "To get this guy [an expression] we move this guy [some term] over there and combine with that guy [another term] to form this guy [a combined term]. And then this guy [another expression] reduces over this guy [a field] because this guy here [a number] is divisible by this guy [the characteristic of the field]." This is accompanied by furious circling (by chalk or finger) and arrow drawing... – Willie Wong – 2010-05-07T14:13:17.800

1I had a professor in Germany who used "Kollege" (cognate with English "colleague", but means something more like "fellow", or more informally, "guy") to refer to mathematical objects in exactly the same way. I have no idea how common this usage is, though. – Mark Meckes – 2010-05-08T15:54:33.517

1@Nate: to be clear, I understand why people might use a word like "guy." It's just always struck me as interesting that mathematicians (at least the ones I talk to) seem to only use that word. I suspect the habit has been passed on from one mathematical generation to the next. – Qiaochu Yuan – 2010-05-17T06:25:16.703

12In the only lecture I saw by David Goss he started with "guy", quickly went to something like "uncanny fellow" and then stayed with "sucker" for most of the talk. I don't know what those poor Drinfeld modules had done to him the day before :-) – Peter Arndt – 2010-05-19T12:24:36.573


Let $V$ be a vector space. Then the intersection of $n$ hyperplanes (i.e. subspaces of codimension 1) is a subspace of codimension at most $n$.

So, naturally, the intersection of countably many hyperplanes is a subspace of countable codimension. Hence if $V$ is of uncountable dimension, this intersection is non-trivial.

Except this is of course wrong. For example, consider the vector space $V:=\mathbb{K}^{\mathbb{Z}}:=\{ f:\mathbb{Z}\to\mathbb{K} \}$ of uncountable dimension. The kernels of the projections $\pi_i:V\to\mathbb{K},\ f\mapsto f(i)$ are hyperplanes. Their intersection is the trivial subspace of $V$, and thus has uncountable codimension.

Max Horn

Posted 2010-05-04T21:02:58.510

Reputation: 3 248

Is it clear that $\mathbb{K}^{\mathbb{Z}}$ is of uncountable dimension for all $\mathbb{K}$ ? (for uncountable $\mathbb{K}$ (like $\mathbb{R}$ or $\mathbb{C}$) one can consider the functions $\frac{1}{X-a}$ with $a\in \mathbb{K}^\times$ which are independant, identifying $\mathbb{K}^{\mathbb{N}}$ with $\mathbb{K}[[X]]$) – Duchamp Gérard H. E. – 2017-10-29T20:59:01.117

OK, you're right, I found the "Erdos-Kaplansky theorem" in Bourbaki (Algebra II § 7 ex. 3) which explains that $dim(\mathbb{K}^J)$ ($J$ infinite) is $card(\mathbb{K})^{card(J)}$. – Duchamp Gérard H. E. – 2017-10-30T07:47:45.817


The following false belief enjoyed a certain success in the '70. (See R.S.Palais, Critical point theory and the minimax principle for an account.)

A second countable, Hausdorff, Banach manifold is paracompact.

Regular is necessary, otherwise there are counterexamples!

Pietro Majer

Posted 2010-05-04T21:02:58.510

Reputation: 35 538


By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

Pierre-Yves Gaillard

Posted 2010-05-04T21:02:58.510

Reputation: 1 435

I don't see how anybody could use language such that (3) is true and (4) is false. After that, it is up to how the terms are defined, but (of course!) I agree that (5) is the way to go here. – Toby Bartels – 2011-04-04T09:13:55.617

At least, ${0}$ is a vector space. I have seen "a vector space has at least two elements" from a professional mathematician. – user11235 – 2011-04-10T21:28:44.147

1I feel like there are a lot of areas in mathematics in which the empty set is interpreted in a certain way (for example, the empty product is one, the empty sum is zero, the empty set has one map into any non-empty set, etc). Given each of these particular situations locally, I might agree that it is a convention in each case. However, given the ubiquity of such "conventions," one might think that there is a uniform description of what the empty set really "means" in these contexts. If this becomes the case, then I might argue for (5), which would follow from this conception of the empty set. – David Corwin – 2010-07-07T23:48:28.727

Given that the free space on the empty set is the zero space (high-fallutin general nonsense-maximizing proof: free-ification is a left adjoint => it is cocontinuous => takes initials to initials + the initial vector space is the zero space and the initial set is the empty set), and that for free spaces $F(X)$, $X$ is a basis, I would definitely say (4) and (5). – G. Rodrigues – 2010-07-22T13:39:20.127

I think one can chase the controversy here down a little further, to the statement: "the sum of the empty set is 0". I think most people who accept this then accept (5). – Peter LeFanu Lumsdaine – 2010-09-27T02:59:05.323


Here's one from basic set theory. Let k be a cardinal and consider the operation "adding k", meaning

$l \mapsto k+l$

on cardinals. We know that this operation "stabilizes" to the identity after $k$, that is, for any $l>k$, we have $l+k = l$. Similarly, the "multiplying by $k$" operation,

$l \mapsto l * k$

stabilizes to the identity after $k$.

Everyone also knows that if $l$ is an infinite cardinal then $l^2$ is equipotent to $l$, and more generally $l^n$ is equipotent to $l$ for every natural number $n$. I.e. all the finite power functions stabilize to the identity at $\omega$.

Well, obviously "exponentiation by $\omega$" also stabilizes at some point, right? Like, $l^\omega$ is equal to $l$ for sufficiently large $l$? Look, we probably already have the stabilization point at $2^\omega$.


Pietro KC

Posted 2010-05-04T21:02:58.510

Reputation: 719

Actually, you can find that belief proclaimed here at MO, until someone points out the mistake. – Todd Trimble – 2015-09-06T02:09:41.077

2Why not? As an algebraist, my reaction already after "addition of k stabilizes" would be "if THAT holds, than WHATEVER". – Victor Protsak – 2010-06-10T06:45:49.530

2Victor, I held this belief for a good while when first learning set theory. I tried proving it a couple of times and failed, but I was in that stage just after I'd gotten the hang of basic cardinality arguments and they all seemed simple, so I figured it was just a matter of small details. – Pietro KC – 2010-06-10T09:01:33.340

6But it turns out that k^l is intimately linked with the cofinality of k, which is the length of the shortest unbounded sequence in k. For example, cof(omega) = omega, since sequences of length less than omega are finite, and thus bounded in omega. Similarly, cof(aleph_1) is aleph_1, since any countable sequence in aleph_1 is bounded. It's not immediately obvious that some cardinal k has cof(k) < k, but aleph_omega does! Anyway, the relevant theorem is that k^cof(k) > k, so there are arbitrarily large k s.t. k^omega > k. – Pietro KC – 2010-06-10T09:06:05.360


Piggybacking on one of Pierre's answers, I once had to teach beginning linear algebra from a textbook wherein the authors at one point stated words to the effect that the the trivial vector space {0} has no basis, or that the notion of basis for the trivial vector space makes no sense. It is bad enough as a student to generate one's own false beliefs without having textbooks presenting falsehoods as facts.

My personal belief is that the authors of this text actually know better, but they don't believe that their students can handle the truth, or perhaps that it is too much work or too time-consuming on the part of the instructor to explain such points. Whatever their motivation was, I cannot countenance such rationalizations. I told the students that the textbook was just plain wrong.

Kurt Luoto

Posted 2010-05-04T21:02:58.510

Reputation: 181

6Thierry: of course it makes sense. But the action is not transitive. – ACL – 2010-12-01T22:53:19.067

@ACL Of course it is transitive - given any x,y in the empty set, the identity element sends x to y. It is even n-transitive for all n! – kow – 2010-12-14T18:05:10.537


@kow: I disagree. That is the "wrong" definition of transitivity for empty G-sets. See the discussion at http://qchu.wordpress.com/2010/12/03/empty-sets/ .

– Qiaochu Yuan – 2010-12-16T23:08:01.253


Probably many mathematicians in the twentieth century didn't accept that the empty set "existed", including for example R.L. Moore. And to this day, a lot of people have trouble comfortably dealing with it, as witnessed in the WP article http://en.wikipedia.org/wiki/First-order_logic#Empty_domains

– Todd Trimble – 2011-03-31T14:26:01.320

@ Todd: That's just because logicians don't know mathematics. – Toby Bartels – 2011-04-04T09:46:26.397

congratulations if you convinced them the empty set is a (minimal) spanning set? – roy smith – 2011-05-10T04:03:16.973

4I once taught abstract algebra from a book that adopted the artificial convention that the domain of a map of sets must be nonempty. I eventually figured out that the reason was in order to be able to say that every one-to-one map has a left inverse. And I have many times taught topology from a book that adopts the artificial convention that when speaking of the product of two spaces we require both spaces to be nonempty. I eventually figured out that the reason was in order to be able to say that $X\times Y$ is compact if and only if both $X$ and $Y$ are compact. – Tom Goodwillie – 2012-03-14T22:01:40.953

My first confusion about empty sets came from Seminaire Bourbaki expose on Macdonald polynomials that established that the sum of a certain empty set is $\pi^2/6$. On the same page the author added that those who have a problem with the above result belong in a kindergarten, which only compounded to my feeling of inadequacy. – Michael – 2013-12-03T01:11:40.780

@Michael Wasn't it the sum of the empty set of half-roots that was equal to $1/6$? This is at least what Demazure writes after writing Kac's character formula for the group $U(1)$. But his witty addition that “this would have unexpected consequences, especially regarding the teaching of mathematics in kindergarten” should not be have been taken seriously! – ACL – 2016-04-21T14:37:54.897

4Bjorn Poonen once gave a lecture at MIT about the empty set; it really opened my eyes. If someone wrote a textbook or something on the matter I think everyone would be a lot less confused. – Qiaochu Yuan – 2010-07-07T23:56:23.400

3For most of the history of civilization, zero was very controversial... – Victor Protsak – 2010-07-09T04:12:32.587

8I can combine Qiaochu's and Victor's remarks in this memory I have of a coffee break conversation between two colleagues, who were arguing on whether it made sense to say that the 1-element group acts on the empty set. I wisely decided to stay out of the controversy... – Thierry Zell – 2010-08-31T02:24:35.240


There is a bijection between the set of [true: prime!] ideals of $S^{-1}R$ and the set of [true: prime!] ideals of $R$ which do not intersect $S$.


Posted 2010-05-04T21:02:58.510

Reputation: 9 845

1Yes! In a review of a text on commutative algebra I have suggested to extend the prime ideal correspondence in localizations to some ideal correspondence, because I wasn't aware that we have to actually use the prime ideal condition somewhere ... – Martin Brandenburg – 2011-04-12T08:41:53.477


" Every open dense subset of $\mathbb{R}^n$ has full Lebesgue measure. "

Selim G

Posted 2010-05-04T21:02:58.510

Reputation: 1 101

(See Craig's answer below) – Pietro Majer – 2013-12-09T12:13:27.943

Well, although similar to Craig's answer on an open neighborhood of $\mathbb{Q}$ with arbitrarily small measure, I find this formulation much more appealing. – Dirk – 2014-01-02T17:18:29.933


Yet another one:

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable. If $f'(x_0) > 0$, then there exists an interval $I$ containing $x_0$ such that $f$ is increasing in $I$.


Posted 2010-05-04T21:02:58.510

Reputation: 1 157

1I sort of find it hard to believe that amongst the nearly 200 answers on this thread (and just over 20 deleted ones), no one has posted this. – Asaf Karagila – 2015-08-10T06:07:44.560

2A counter-example is necessarily with $f'$ discontinuous in $x_0$, right? For example $f(x)=x^2 sin (1/x)+x/2$ and $x_0 = 0$. – Sebastien Palcoux – 2015-08-10T08:05:33.080

1@SébastienPalcoux Yes, I think if $f'$ is continuous in $x_0$ then the statement is true. – Shamisen – 2015-08-10T15:16:30.497


This one has bit me and some very good mathematicians I know.

Let $X,Y$ be Banach spaces, and let $E \subset X$ be a dense subspace. Suppose $T : E \to Y$ is a bounded linear operator. Then $T$ has a unique bounded extension $\tilde{T} : X \to Y$. (True, this is the well-known and elementary "BLT theorem".)

If $T$ is injective then so is $\tilde{T}$. (False! See this answer for a counterexample.)

Nate Eldredge

Posted 2010-05-04T21:02:58.510

Reputation: 17 394


A common belief of students in real analysis is that if $$ \lim_{x\to x_0}f(x,y_0),\qquad\lim_{y\to y_0}f(x_0,y) $$ exist and are both equal to $l$, then the function has limit $l$ in $(x_0,y_0)$. It is easly to show counter-examples. More difficult is to show that also the belief $$ \lim_{t\to 0}f(x_0+ht,y_0+kt)=l,\quad\forall\;(h,k)\neq(0,0)\quad\Rightarrow\quad\lim_{(x,y)\to(x_0,y_0)}f(x,y)=l $$ is false. For completeness's sake (presumably anybody who ever taught calculus has seen it, but it's easily forgotten) the standard counterexample is $$ f(x,y)=\frac{xy^2}{x^2+y^4} $$ at $(0,0$).


Posted 2010-05-04T21:02:58.510

Reputation: 101

3That counterexample has the advantage of being well-behaved away from $(0,0)$, but the (related) disadvantages of being easily forgotten and requiring a bit of thought to come up with. This can make things look trickier than they are. For this reason, I prefer brain-dead counterexamples like $f(x,y)=1$ if $y=x^2 \neq 0$, $f(x,y)=0$ otherwise. – Chris Eagle – 2011-01-12T17:11:07.577

3@Chris As you know, this is not a "real function" to the minds of calculus students. – Ryan Reich – 2014-01-02T03:04:50.247

1Can I try to generate a simpler counterexample? Consider $f(x,y)=\begin{cases}1,&x^2+y^2=1\0,&x^2+y^2\ne1\end{cases}$. Then it's not hard to show that all straight-line limits to $(x_0,y_0)$ exist for all $x_0,y_0$, and are equal to $0$, but clearly the limit doesn't exist on the unit circle. EDIT: Didn't see Eagle's comment. – Akiva Weinberger – 2015-09-01T00:07:30.180


In a finite abelian $p$-group, every cyclic subgroup is contained in a cyclic direct summand.

Added for Gowers: Maybe one reason why people fall into this error goes something like this: First you learn linear algebra, so you know about vector spaces, bases for same, splittings of same. Then you run into elementary abelian $p$-groups and recognize this as a special case of vector spaces. Then you learn the pleasant fact that all finite abelian $p$-groups are direct sums of cyclic $p$-groups, and a corresponding uniqueness statement. You notice that all of the cyclic subgroups of order $p^2$ in $\mathbb Z/p^2\times \mathbb Z/p$ are summands, and if you have a certain sort of inquiring mind then you also notice that not every subgroup of order $p$ is a summand: one of them is contained in a copy of $\mathbb Z/p^2$, in fact in all of those copies of it. Having learned so much, both positive and negative, from the example of $\mathbb Z/p^2\times \mathbb Z/p$, you may think that it shows all the interesting basic features of the general case and overlook the fact that in $\mathbb Z/p^3\times \mathbb Z/p$ there is a $\mathbb Z/p^2$ not contained in any $\mathbb Z/p^3$.

In any case, reputable people sometimes make this blunder; it happened to somebody here at MO just the other day.

Tom Goodwillie

Posted 2010-05-04T21:02:58.510

Reputation: 37 782

Can you sketch the "proof" that makes this plausible? – gowers – 2010-07-07T16:40:53.780

Finite abelian $p$-groups are direct sum of cyclic subgroups so they look a bit like vector spaces. Therefore, you expect them to behave the same way, i.e. every subspace should have a complement. In other words, take a minimal generating set for your subgroup and complete it to a minimal generating set for the whole group.

This fails since your generating set for the subgroup might be depended modulo the Fratinni subgroup of the whole group. (A set is a minimal generating set for a finite $p$-group iff it is abasis for the group modulo the Fratinni subgroup). – Yiftach Barnea – 2010-07-07T17:58:46.620

Is there an easily stated classification of the ways one can place a subgroup inside a finite abelian p-group (up to automorphisms of the larger group)? – T.. – 2010-07-07T22:31:16.963

I once worked out a classification of the ways one can place an element inside a finitely generated abelian group (up to automorphisms of the larger group), but I don't recall how it went exactly. – Tom Goodwillie – 2010-07-08T00:24:12.437

This is related to a somewhat subtle issue of characterizing inclusions between the closures of the conjugacy classes of matrices. Suppose $A$ is a nilpotent $n\times n$ matrix of type $\lambda$ (i.e. with Jordan blocks of sizes $\lambda_1\geq \lambda_2\geq\ldots$ adding up to $n$) and $B$ is ... $\mu.$ Can $B$ be obtained as a limit of the conjugates of $A$? This is clearly possible if $\lambda$ is componentwise greater or equal than $\mu$, but the necessary and sufficient condition is given by the dominance order, http://en.wikipedia.org/wiki/Dominance_order.

– Victor Protsak – 2010-07-09T04:05:32.467


I'm pretty sure I've heard both of the following multiple times:

  1. Transfinite induction requires the axiom of choice. False, though many applications of transfinite induction require axiom of choice (either in the form of the well-ordering theorem, or directly (though using transfinite induction together with choice directly is essentially the same as just using Zorn's Lemma)).

  2. Transfinite induction requires the axiom of foundation. I guess some people get transfinite induction mixed up with epsilon-induction?

Harry Altman

Posted 2010-05-04T21:02:58.510

Reputation: 1 352


The cost of multiplying two $n$-digit numbers is of order $n^2$ (because each digit of the first number has to be multiplied with each digit of the second number).

A lot of information is found on http://en.wikipedia.org/wiki/Multiplication_algorithm .

The first faster (and easily understandable) algorithm was http://en.wikipedia.org/wiki/Karatsuba_algorithm with complexity $n^{log_2 3} \sim n^{1.585}$.

Basic idea: To multiply $x_1x_2$ and $y_1y_2$ where all letters refer to $n/2$-digit parts of $n$-digit numbers, calculate $x_1 \cdot y_1$, $x_2\cdot y_2$ and $(x_1+x_2)\cdot(y_1+y_2)$ and note that this is sufficient to calculate the result with three such products instead of four.


Posted 2010-05-04T21:02:58.510

Reputation: 453

2It would be better if these misconceptions would come with explanations how things really are... – darij grinberg – 2011-04-10T18:28:11.560

1Along these lines: there is a widespread misapprehension that multiplication is the same thing as a multiplication algorithm (whichever one the speaker learned in elementary school). – Thierry Zell – 2011-04-10T19:25:58.123

5At least it's better than people thinking multiplication is constant-time. :P – Harry Altman – 2011-04-10T19:35:19.183

Nice, thanks for the explanation! – darij grinberg – 2011-04-10T20:19:37.107

should be $(x_1+x_2)(y_1+y_2)$. – Junyan Xu – 2012-05-06T04:09:31.827

@Junyan Xu : Thanks, I corrected it. – user11235 – 2012-05-09T18:18:13.947


It took me a bit too long to realize that these two beliefs are contradictory:

  • Period 3 $\Rightarrow$ chaos: if a continuous self-map on the interval has a period-3 orbit, then it has orbits of all periods.
  • The black dots on each horizontal slice of this picture above $x=a$ show the location of the periodic points of the logistic map $f_a(y) = ay(1-y)$: Bifurcation diagram for the logistic map

You can clearly see a 3-cycle in the light area towards the right; yet we know that if there is a 3-cycle in that slice then there must be a cycle of any period in that slice... so where are they?

(The other cycles are there of course, but they are repelling and hence are not visible. You can see artifacts from these repelling cycles near the period-doubling bifurcations in this picture)

Matt Noonan

Posted 2010-05-04T21:02:58.510

Reputation: 2 026


Here's a relevant back-and-forth about the use of the term 'chaos' in the AMS Notices (in response to Freeman Dyson's 'Birds and Frogs'):



– Mike Hall – 2011-08-08T00:17:26.643


Let $(X,\tau)$ be a topological space. The false belief is: "Every sequence $(x_n)$ in $X$ with an accumulation point $a\in X$ has a subsequence that converges to $a$". I subscribed to this intuitively until I stumbled over a counterexample, see http://dominiczypen.wordpress.com/2014/10/13/accumulation-without-converging-subsequence/

Dominic van der Zypen

Posted 2010-05-04T21:02:58.510

Reputation: 10 887


I confess that I didn't carefully comb through all the answers, although I've read through this thread a few times in the past. So maybe these are repeats.

  • "The category of compact Hausdorff spaces is complete but not cocomplete; for example, it doesn't have all coproducts."

  • "The category of torsion abelian groups is cocomplete but not complete; for example, it doesn't have all products."

One of my professors in graduate school (quite a well-known and strong mathematician actually) insisted on the first, and quite a few people here at MO have mistakenly believed the second before the error was pointed out.

The moral of the story: sometimes categorical limits/colimits aren't computed the way you might first think of, e.g., colimits of compact Hausdorff spaces aren't always computed as colimits in $\mathrm{Top}$, and limits of torsion abelian groups aren't always computed as limits in $\mathrm{Ab}$.

Todd Trimble

Posted 2010-05-04T21:02:58.510

Reputation: 41 195


Sequence $\{a_n\}$ has a limit $A$ in $\mathbb{R}$ and a limit $B$ in $\mathbb{Q}_p$. Then $A$ is rational iff $B$ is rational.

Alexey Ustinov

Posted 2010-05-04T21:02:58.510

Reputation: 6 017

3Or: if a sequence has a rational limit in Q_p and in Q_r, then they're the same. – Qiaochu Yuan – 2010-05-05T04:16:51.647

1But if a rational sequence has a limit in all Qp, including Q\infty ... – Gerald Edgar – 2010-05-05T12:17:51.633


"A 'random' number field has large class number"

I've heard this belief quite a few times. Usually random means taking a not-too-small degree (7?) and then somehow taking integer coefficients (around 10,000?).

But in fact class number tend to be much smaller than one expects. Usually they are logarithmic in the size of the discriminant.

The main reasons for the belief are the common examples of fields given in undergraduate and early graduate courses - imaginary quadratic fields and cyclotomic fields. In more advanced courses students see abelian extensions and CM-fields, which also have special arithmetic properties that make their class groups somewhat larger. In the courses I have taken the actual size of 'random' number fields was not addressed, and, say, the Cohen-Lenstra heuristics were not mentioned.

Dror Speiser

Posted 2010-05-04T21:02:58.510

Reputation: 1 864


I got 2 well earned downvotes for a false belief I claimed proudly, it is time to balance that by exposing it here:

Let $(P,\le)$ be any poset, and let $\le^*$ be an order on $P$ extending $\le$. Any Endomorphism of $\le^*$ also is an endomorphism of $\le$

($f:P\to P$ endomorphism of $\le$ meaning $x\le y \implies f(x)\le f(y)$).

Of course this is a particular case of a very general fallacy: by extending $\le$ into $\le^*$ one weakens both the conclusion and the premise of the implication, so that there is no general relation between orders that extend one another.

Benoît Kloeckner

Posted 2010-05-04T21:02:58.510

Reputation: 10 345


I don't know how common this is, but I've noticed it half an hour ago in some notes I had written: If $J$ is a finitely generated right ideal of a not necessarily commutative ring $R$, and $n$ is natural, then $J^n$ is finitely generated, isn't it?

No, it isn't. For an example, try $R=\mathbb Z\left\langle X_1,X_2,X_3,...\right\rangle $ (ring of noncommutative polynomials) and $J=X_1R$.

darij grinberg

Posted 2010-05-04T21:02:58.510

Reputation: 15 489

Omg, I will have to be careful about that. Thanks Darij ;). – Martin Brandenburg – 2011-04-12T08:45:41.523


False statement: If $A$ and $B$ are subsets of $\mathbb{R}^d$, then their Hausdorff dimension $\dim_H$ satisfies

$$\dim_H(A \times B) = \dim_H(A) + \dim_H(B). $$

EDIT: To answer Benoit's question, I do not know about a simple counterexample for $d = 1$, but here is the usual one (taken from Falconer's "The Geometry of Fractal Sets"):

Let $(m_i)$ be a sequence of rapidly increasing integers (say $m_{i+1} > m_i^i$). Let $A \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_j + 1 \leq r \leq m_{j+1}$ and $j$ is odd. Let $B \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_{j} + 1 \leq r \leq m_{j+1}$ and $j$ is even. Then $\dim_H(A) = \dim_B(A) = 0$. To see this, you can cover $A$, for example, by $10^k$ covers of length $10^{- m_{2j}}$, where $k = (m_1 - m_0) + (m_3 - m_2) + \dots + (m_{2j - 1} - m_{2j - 2})$.

Furthermore, if $\mathcal{H}^1$ denotes the Hausdorff $1$-dimensional (metric) outer measure of $E$, then the result follows by showing $\mathcal{H}^1(A \times B) > 0$. This is accomplished by considering $u \in [0,1]$ and writing $u = x + y$, where $x \in A$ and $y \in B$. Let $proj$ denote orthogonal projection from the plane to $L$, the line $y = x$. Then $proj(x,y)$ is the point of $L$ with distance $2^{-1/2}(x+y)$ from the origin. Thus, $proj( A \times B)$ is a subinterval of $L$ of length $2^{-1/2}$. Finally, it follows:

$$ \mathcal{H}^1(A \times B) \geq \mathcal{H}^1(proj(A \times B)) = 2^{-1/2} > 0. $$


Posted 2010-05-04T21:02:58.510

Reputation: 101

1Well, it's disappointing that this fails, although it hadn't occurred to me to conjecture it. – Toby Bartels – 2011-04-04T09:53:04.497

2Actually, the situation is worse than I say: there exist sets $A, B \subset \mathbb{R}$ with $dim_H(A \times B )= 1$, and yet $\dim_h(A) = \dim_H(B) = 0$. – JavaMan – 2011-04-05T06:22:27.343

By the way, is there a simple counter-example with $A=B$? – Benoît Kloeckner – 2011-05-09T07:51:19.193

2Nice, I did not know that, though Hausdorff dimension is part of my mathematical life! But the sets I study (Julia sets in complex dimension one) usually are uniform enough that this does not occurr, I guess. Here's what happens, morally, in the example given here: the scales epsilon at which you have good covers of A and the scales at which you have good covers of B are disjoint. The products of these good covers are extremely distorted : they are thin rectangles, instead of squares. – Arnaud Chéritat – 2015-10-18T13:25:15.487


(*) "Let $(I,\leq)$ be a directed ordered set, and $E=(f_{ij}:E_i\to E_j)_{i\geq j}$ be an inverse system of nonempty sets with surjective transition maps. Then the inverse limit $\varprojlim_I\,E$ is nonempty."

This is true if $I=\mathbb{N}$ ("dependent choices"), and hence more generally if $I$ has a countable cofinal subset. But surprisingly (to me), those are the only sets $I$ for which (*) holds for every system $E$. (This is proved somewhere in Bourbaki's exercises, for instance).

Of course, other useful cases where (*) holds are when the $E_i$'s are finite, or more generally compact spaces with continuous transition maps.

Laurent Moret-Bailly

Posted 2010-05-04T21:02:58.510

Reputation: 12 501


"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)

Richard Borcherds

Posted 2010-05-04T21:02:58.510

Reputation: 16 053


A common trap which sometimes I see people fall is that a Hermitian matrix $M$ is negative definite if and only if its leading principal minors are negative.

What is true is the Sylvester's criterion, which says that $M$ is positive definite if and only if its principal minors are positive. Thus, the true statement is that $M$ is negative definite if and only if the principal minors of $-M$ are positive.


Posted 2010-05-04T21:02:58.510

Reputation: 1 157


I guess you don't want commonly held beliefs of students that for every real number there is a next real number, or that convergent sequences are eventually constant. A version I saw in a book asked whether points on a line "touch." Understanding the topology of a line is a challenge for many students, although presumably not for most mathematicians.

Here is a more esoteric belief that I have even seen in some books:

"The Banach-Tarski Paradox says that a ball the size of a pea can be cut into 5 pieces and reassembled to make a ball the size of the sun."

As a consequence of the Banach-Tarski paradox, a ball the size of a pea can be partitioned (not really "cut") into a finite number of pieces which can be reassembled into a ball the size of the sun, but a simple outer measure argument implies that the number of pieces must be very large (I roughly estimate at least $10^{30}$). The number 5 probably comes from the fact that the basic Banach-Tarski paradox is that a ball of radius 1 can be partitioned into 5 pieces which can be reassembled into two disjoint balls of radius 1. (It can almost, but not quite, be done with four pieces; one of the five pieces can be taken to be a single point.)

Bruce Blackadar

Posted 2010-05-04T21:02:58.510

Reputation: 770


Since points do not touch, this was an objection to the set theoretic view of the geometric continuum as a set of points, for example by Veronese. A decent account of this can be found in Debates about infinity in mathematics around 1890: The Cantor-Veronese controversy, its origins and its outcome, by Detlef Laugwitz.

– Andrés E. Caicedo – 2013-12-02T23:22:39.267

Related. – Andrés E. Caicedo – 2014-01-02T04:03:49.527

Convergent sequences are eventually constant! With the discrete topology/metric/norm, that is. – Akiva Weinberger – 2015-09-01T00:19:34.173

Of course I meant sequences in $\mathbb{R}$ with the usual topology. Hopefully by the time students study general metric spaces or topological spaces they understand the topology of $\mathbb{R}$. – Bruce Blackadar – 2016-01-07T19:35:07.853


If $H$ and $K$ are subgroups of $G$, then $HK$ is a subgroup of $G$.


Posted 2010-05-04T21:02:58.510


7Hm, wonder how common that false belief actually is. It seems obviously implausible in the nonabelian case. – Todd Trimble – 2015-09-06T15:39:32.880

1Its common, specially when undergraduates use product formula : $|HK|=\frac{|H||K|}{| H \cap K | }$ Because all of them are subgroup, except $HK$ probably. – None – 2016-04-10T17:30:37.443


Some things from pseudo-Riemannian geometry are a bit hard to swallow for students who have had previous exposure to Riemannian geometry. Aside from the usual ones arising from sign issues (like, in a two dimensional Lorentzian manifold with positive scalar curvature, time-like geodesics will not have conjugate points), an example is that in Riemannian manifolds, connectedness + geodesic completeness implies geodesic connectedness (every two points is connected by a geodesic). This is not true for Lorentzian manifolds, and the usual example is the pseudo-sphere.

Willie Wong

Posted 2010-05-04T21:02:58.510

Reputation: 16 649


I just realized yesterday that, given $A \to C, B \to C$ in an abelian category, the kernel of $A \oplus B \to C$ is not the direct sum of the kernels of $A \to C, B \to C$.

Akhil Mathew

Posted 2010-05-04T21:02:58.510

Reputation: 12 685


"the quadratic variation of a Brownian motion between $0$ and $T$ is equal to $T$"

this is only true that if $\mathcal{D}^N$ is a nested sequence of partitions of $[0,T]$ (with mesh size going to $0$) then the quadratic variation of a Brownian motion along these partitions converges towards $T$, almost surely. If we define the quadratic variation of a continuous function $f$ as we would like to, $$Q(f,[0,T]) = \sup_{0=t_0<\ldots, t_n=T } \sum |f(t_k)-f(t_{k+1})|^2,$$ then the Brownian paths have almost surely infinite quadratic variation.

This was something I had never noticed until I read the wonderful book "Brownian motion" by Peter Morters and Yuval Peres.


Posted 2010-05-04T21:02:58.510

Reputation: 1 378

1The key here is that quadratic variation is defined as a limit in probability, not a limit almost surely. – nullUser – 2013-07-08T15:46:06.327


Complex variables: "An entire function that is onto and locally one-to-one is globally one-to-one."

Counterexample: $f(z) := \int_0^z \exp(\zeta^2)\,d\zeta$

I'll leave the proof that this is indeed a counterexample as a pleasant exercise.

(I believe this example is due to Lawrence Zalcman.)

Daniel Asimov

Posted 2010-05-04T21:02:58.510

Reputation: 856

1@MichaelHardy, if we're going to {\TeX}pick, then surely it should be something like ${\mathrm d}\zeta$ (rather than $d\zeta$), since the $\mathrm d$ is an operator (rather than a variable)? – LSpice – 2013-12-12T23:20:00.830

1@LSpice : I understand the case for that usage; in particular, it allow the use of $d$ as a variable, so that one can write $\dfrac{\mathrm{d}f}{\mathrm{d}d}$, etc. However, the usage with the $d$ italicized as if it were a variable is standard although not universal. – Michael Hardy – 2013-12-13T00:58:36.783

Let's see if you TeX code can be improved: $$ f(z) := \int_0^z \exp(\zeta^2)\,d\zeta $$ (The backslash in \exp not only should prevent italicization but should also result in proper spacing in things like "a \exp b", and the space before d\zeta seems appropriate.) – Michael Hardy – 2010-07-08T15:19:27.137


False belief: A function being continuous in some open interval implies that it is also differentiable on some point in that interval:


The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere:

$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$

Where $a \in (0, 1)$, $b$ is a positive odd integer, and $ab > 1 + \frac{3\pi}{2}$. The function has fractal-like behavior, which leads to it not being differentiable. This notion is rather disheartening to most calculus students, though.

Jon Paprocki

Posted 2010-05-04T21:02:58.510

Reputation: 314

Did you mean "differentiable on some point in that interval:" ? – Rasmus – 2013-09-18T19:13:20.053

Haha, figures that I edit a 3 year old post to introduce an even worse typo. Yes, that is what I meant. – Jon Paprocki – 2013-09-18T22:53:26.443

3Related: if f is continuous on the interval I, there must be an interval J in I on which f is monotone. Easily believed by the beginner. – Thierry Zell – 2010-08-31T02:34:27.893


The fundamental group of the Klein bottle is $D_\infty$, the infinite dihedral group (which is $\mathbb Z \rtimes \mathbb Z_2$).

I believed this for some time, and I seem to recall some others having the same confusion.

The group that has been mistaken for $D_\infty$ is in fact $\mathbb Z \rtimes\mathbb Z$, which can also be written with the presentation $x^2y^2=1$. The former abelianizes to $\mathbb Z_2\oplus \mathbb Z_2$, the latter to $\mathbb Z\oplus \mathbb Z_2$.

A 2-dimensional Lie group is a product of circles and lines, in particular it is abelian.

I don't know if anyone else suffered this one. The mistake is (a) in forgetting that the classification of surfaces doesn't apply since homeomorphic Lie groups are not necessarily isomorphic (e.g., the (bijective, orientation preserving) affine transformations $x\mapsto ax+b$, where $a>0, b\in \mathbb R$ are homeomorphic to $\mathbb R^2$, though not isomorphic) and (b) that Lie groups aren't necessarily connected, in particular $\mathbb R^2$ cross any finite non-abelian group is non-abelian.


Posted 2010-05-04T21:02:58.510

Reputation: 260

Count me in for the 2nd fallacy. – Michael – 2013-12-03T00:41:42.727


As is well known, if $V$ is a vector space and $S, T \subset V$ are subspaces, then $S \cup T$ is a subspace iff $S \subset T$ or viceversa. However, $S \cup T \cup U$ can be a subspace even if no two spaces are contained in each other (think finite fields...)


Posted 2010-05-04T21:02:58.510

Reputation: 1

6But only finite fields... – darij grinberg – 2010-10-19T08:42:53.140


A degree $k$ map $S^n\to S^n$ induces multiplication by $k$ on all the homotopy groups $\pi_m(S^n)$.

(Not sure if this is a common error, but I believed it implicitly for a while and it confused me about some things. If you unravel what degree $k$ means and what multiplication by $k$ in $\pi_m$ means, there's no reason at all to expect this to be true, and indeed it is false in general. It is true in the stable range, since $S^n$ looks like $\Omega S^{n+1}$ in the stable range, "degree k" can be defined in terms of the H-space structure on $\Omega S^{n+1}$, and an Eckmann-Hilton argument applies.)

Eric Wofsey

Posted 2010-05-04T21:02:58.510

Reputation: 26 213

3If $n$ is even and $x \in \pi{2n-1}(S^n)$ and $f$ a degree $k$ map and $H$ the Hopf invariant, then $H(f* (x)) = k^2 H(x)$. A related misbelief: if $M$ is a framed manifold and $N\to $M a finite cover, of degree $d$. Then the framed bordism classes satisfy $[N]=d [M]$. Completely wrong. – Johannes Ebert – 2011-04-14T09:04:34.417


Here are mistakes I find surprisingly sharp people make about the weak$^{*}$ topology on the dual of $X,$ where $X$ is a Banach space.

-It is metrizable if $X$ is separable.

-It is locally compact by Banach-Alaoglu.

-The statement $X$ is weak$^{*}$ dense in the double dual of $X$ proves that the unit ball of $X$ is weak$^{*}$ dense in the unit ball of the double dual of $X.$

The first two are in fact never true if $X$ is infinite dimensional. While both statements in the third claim are true, the second one is significantly stronger, but a lot of people believe you can get it from the first by just "rescaling the elements" to have norm $\leq 1.$ (Although the proof of the statements in the third claim is not hard). The difficulty is that if $X$ is infinite dimensional then for any $\phi$ in the dual of $X,$ there exists a net $\phi_{i}$ in the dual of $X$ with $\|\phi_{i}\|\to \infty$ and $\phi_{i}\to \phi$ weak$^{*},$ so this rescaling trick cannot be uniformly applied. Really these all boil down to the following false belief:

-The dual of $X$ has a non-empty norm bounded weak$^{*}$ open set.

Again when $X$ is infinite dimensional this always fails.

Benjamin Hayes

Posted 2010-05-04T21:02:58.510

Reputation: 221

Minor nitpick: Consider a locally compact Hausdorff space $T$. The $$ topology on the dual of the $C^$ algebra $C_0(T)$ is metrizable, if and only if $X$ is second countable. That is a theorem in Choquet's book on functional analysis. So your claim, that the first statement is never true in infinite dimensional situations, is false. Take e.g. $T$ being a circle. – Marc Palm – 2011-10-06T13:38:44.597

1I think $M(T)$ is not metrizable in the weak$^\ast$ topology, and in fact my claim that this fails for every infinite dimensional Banach space i also think is true. The rough outline of the proof I saw was this:

  1. If $X^\ast$ is weak$^\ast$ metrizable, then a first countabliity at the origin argument implies that $X^\ast$ has a translation invariant metric given the weak$^\ast$ topology.
  2. One can characterize completeness topologically for translation-invariant metrics, and see directly that if $X^\ast$ had a translation-invariant metric given the weak$^\ast$ topology it would be complete.
  3. < – Benjamin Hayes – 2011-10-12T03:42:27.390

$X^{∗}$ in the weak∗ topology is a countable union of ${\phi\in X^{*}:|\phi|\leq N}$, which have empty weak∗ interior. Hence, if the weak∗ topology were metrizable, we get a contradiction to the Baire Category Theorem. Are you sure you don't mean the weak∗ topology on the state space of $C_{0}(X)? – Benjamin Hayes – 2011-10-12T03:47:20.890

Okay, excuse my false claim, I was overlooking that this holds for the subset $M^+(T)$ of positive Radon measure, and does not generalize to the complex linear span. – Marc Palm – 2011-10-16T10:24:01.700


A random $k$-coloring of the vertices of a graph $G$ is more likely to be proper than a random $(k-1)$-coloring of the same graph.

(A vertex coloring is proper if no two adjacent vertices are colored identically. In this case, random means uniform among all colorings, or equivalently, that each vertex is i.i.d. colored uniformly from the space of colors.)


Posted 2010-05-04T21:02:58.510

Reputation: 2 361

2...wait, what's the truth then? – Harry Altman – 2011-05-10T00:06:09.620

It sounds plausible. – Michael Hardy – 2011-05-10T00:34:59.980

3For some graphs $G$ and integers $k$, the opposite. The easiest example is the complete bipartite graph $K{n,n}$ with $k=3$. The probability a $2$-coloring is proper is about $(1/4)^n$ while the same for a $3$-coloring is about $(2/9)^n$, where I've ignored minor terms like constants. The actual probabilities cross at $n=10$, so as an explicit example, a random $2$-coloring of $K{10,10}$ is more likely to be proper than a random $3$-coloring. – aorq – 2011-05-10T00:37:16.903

7This seems like a good example of a counterintuitive statement, but to call it a common false belief would mean that there are lots of people who think it's true. The question would probably never have occurred to me it I hadn't seen it here. The false belief that Euclid's proof of the infinitude of primes, on the other hand, actually gets asserted in print by mathematicians---in some cases good ones. – Michael Hardy – 2011-05-10T15:36:27.127


Common false belief: a space that is locally homeomorphic to $\mathbb{R}^n$ must be Hausdorff. More generally, many people forget that the usual definition of a manifold contains the Hausdorff and paracompact conditions.

There are of course examples that show that forgetting this assumption leads to unexpected result, and they are in fact much wilder than I knew a few weeks ago. Notably, among examples of (Hausdorff) non-paracompact "manifolds" are the well-known long line, but also the Prüfer manifold constructed from a closed half-plane by attaching to it a half plane at each boundary point.

Added: Let me give a particular case of this false belief to illustrate what kind of weird things can happen that most people would not realize when they are sloppy with the paracompact hypothesis: there exists a path-connected, locally contractible, simply-connected space that admits non-trivial locally trivial bundles with fiber $[0,1]$. Indeed, the first octant in the product of two long line is not homeomorphic to a product a long ray with an interval, but has a natural bundle structure over a long ray.

Benoît Kloeckner

Posted 2010-05-04T21:02:58.510

Reputation: 10 345


Teaching introduction to analysis, I had students using the "fact" that if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $[a,b]$ can be divided to subintervals $[a,c_1],[c_1,c_2],...,[c_n,b]$ such that $f$ is monotone on every subinterval. For instance you can use this "fact" to "prove" the (true) fact that $f$ must be bounded on $[a,b]$. Also, some students used the same "fact", but with countably many subintervals. I found this mistake hard to explain to students, because constructing a counterexample (such as the Weierstrass function) is impossible at the knowledge level of an introduction course.

Izhar Oppenheim

Posted 2010-05-04T21:02:58.510

Reputation: 455

3Why not $x \sin(1/x)$ as example? – None – 2014-01-02T17:33:50.993

1It is in the case of finitely many subintervals, but not in the case of countably many subintervals. – Izhar Oppenheim – 2014-01-02T19:17:04.310

2You can surely discuss fractal shapes without needing to go into the details of a technical counterexample. The point seems to be that it is hard to imagine that "increasing at a point" and "increasing in a neighborhood of a point" are not the same for continuous functions. You can give easy examples showing that indeed they disagree, locally, and fractals suggest that you can make the disagreement happen everywhere. You can revisit this later, once more technology has been set in place. – Andrés E. Caicedo – 2014-01-02T23:44:51.143

While technically it is true one can do it with countably many for the function I gave (if one includes degenerate intervals) I would be surprised if not at least some (or rather most) of the confusion of the students could be addressed by the example (possibly continuing with discussion along the lines suggested by @AndresCaicedo). – None – 2014-01-05T16:50:06.143


The quaternions $\{x+yi+zj+wk\mid x,y,z,w\in \mathbb{R}$} is a complex Banach algebra (with usual operations). Hence it is apparently a counterexample to the Gelfand-Mazur theorem

So, what is the error?

The error is the following:

However the quaternion is a vector space over the field of complex number and it is also a ring, but there is no compatibility between scalar multiplication and quaternion multiplication). So it is not a complex algebra. This shows that in the definition of a complex algebra $A$, the commutative condition $\lambda (ab)=(a)(\lambda b),\;\;\lambda \in \mathbb{C},\;\;a,b\in A$, is very essential.

Ali Taghavi

Posted 2010-05-04T21:02:58.510

Reputation: 234

4This is not a common false belief except among people who do not understand the definition of an algebra over a field – Yemon Choi – 2014-11-12T23:43:53.040

2Moreover, surely the quaternions are a real vector space, not a complex vector space – Yemon Choi – 2014-11-13T01:34:11.440

@YemonChoi the field of complex number is a subring of the ring of quaternions. So quaternions is a complex vec. space.More generaly if a ring R contains a field F then R is a F-vector space,Ok? – Ali Taghavi – 2014-11-13T05:31:16.980

@YemonChoi I think this example is perhaps interesting unless a participant do not read it carefully. please think again to the main motivation and aim of the question of "common false..." – Ali Taghavi – 2014-11-13T05:44:41.497

6@AliTaghavi You're right that $R$-multiplication induces an $F$-module ($F$-vector space) structure via the evident composite $F \times R \to R \times R \to R$. To be fair to both you and Yemon: a very common slip even among professionals is in knowing that for commutative algebras an $F$-algebra is tantamount to a homomorphism $F \to R$, but temporarily forgetting this doesn't apply in the noncommutative setting (except of course when $F$ is central in $R$) -- not rising to the level of false belief so much as a temporary slip-up. I've made that slip myself! – Todd Trimble – 2014-11-13T11:58:23.690

Thanks Ali and thanks @ToddTrimble for the further explanation – Yemon Choi – 2014-11-13T13:29:58.940

@ToddTrimble Thanks for the comment. Some years ago I realy was shocked when I observed that Gelfand mazur theorem is not true! then I realized that it is not an algebra:) – Ali Taghavi – 2014-11-13T15:17:01.400


"A real symmetric matrix is positive-definite iff all the leading principal minors are positive, and positive-semidefinite iff all the leading principal minors are nonnegative."

This paper collects some evidence that this belief is "common", and presents a counterexample (of size $3\times 3$. Exercise: find an example of size $2\times 2$).

(Related to, but not the same as this answer.)

Frieder Ladisch

Posted 2010-05-04T21:02:58.510

Reputation: 4 932

3$$\pmatrix{0&0\cr0&-1\cr}$$ – Gerry Myerson – 2015-07-29T03:22:42.280


False belief: Any orthonormal basis of a subvectorspace $W\subset V$ of an inner product space $V$ can always be extended to an ONB of $V$.

Counterexample: Let $V$ be $\bigoplus_{i\ge 1} \mathbb{R}$ with the inner product given by $\langle a_*,b_*\rangle =\sum_{i\ge 1} a_ib_i$ and let $W$ be the subvectorspace of $V$ spanned by $e_1+e_i$ for $i\ge 2$. The given set is basis and we can apply Gram-Schmidt to obtain an ONB.

However $W^\perp = 0$ so there is no way to complete it. Related false belief: $(W^\perp)^\perp=W$. These beliefs are all true in finite dimensions, but false in general.


Posted 2010-05-04T21:02:58.510

Reputation: 4 554

4That's why we like Hilbert spaces (inner product spaces that are complete w.r.t. the inner-product norm) much better than arbitrary inner-product spaces. – Noam D. Elkies – 2016-03-04T04:08:26.827


An incredibly common false belief is:

For a (say smooth, projective) algebraic variety $X$ the $K_X$-negative part of the cone $NE(X)$ is locally polyhedral.

A right statement of the theorem of the cone is

$\overline{NE(X)} = \overline{NE(X)}_{K_X \geq 0} + \sum_{i} \mathbb{R}[C_i]$ for a denumerable set $\{ C_i \}$ of rational curves, which accumulate at most on the hyperplane $K_X = 0$.

At a first glance this seems to imply that $\overline{NE(X)}_{K_X < 0}$ is locally poyhedral, but this is not true. It depends on the shape of the intersection $\overline{NE(X)} \cap \{ K_X = 0 \}$.

For instance if this latter intersection is round, and there is only one curve $C_i$, the half-cone $\overline{NE(X)}_{K_X < 0}$ is actually a circular cone! Definitely not polyhedral in any sense. I believe this behaviour can happen even with varieties birational to abelian varieties.

The strange thing about this false belief is that it is held true by many competent mathematicians (and indeed I don't believe that many undergraduates meet the theorem of the cone!).

Andrea Ferretti

Posted 2010-05-04T21:02:58.510

Reputation: 8 136

You meant: I believe this behaviour can happen even with (varieties birationally isomorphic to) abelian varieties. Nice example although perhaps too technical for MO. – VA. – 2010-05-05T03:27:01.900

76Incredibly common? The number of people who can even understand the statement, let alone believe it, isn't all that large... – Victor Protsak – 2010-05-05T06:57:40.003

6Yes, but among those, almost all believe that the wrong version is true. – Andrea Ferretti – 2010-05-05T10:13:27.120

7And about 50% of the large community who cannot understand the point will believe that the right version is true! Rather high percentage... – Wadim Zudilin – 2010-05-05T11:41:56.433

I'm not sure to what extent this is a "false belief", and to what extent people are just being sloppy with the terminology "locally polyhedral". But I agree, it's disturbing to hear experts happily making this false statement, without any further comment.

<i>Mea culpa:</i> An old version of the wikipedia article entitled "Cone of curves" contained this false statement. If one looks through the article history, it's not hard to see who is to blame... – None – 2010-05-06T07:24:44.343

Outstanding example. Thank you Andrea. – inkspot – 2010-08-26T14:27:35.070


I had in mind that a $0$-sphere is only one point, but it is false, it is a collection of two points: $$\mathbb{S}^0 = \{ x \in \mathbb{R} \ \ | \ \ \|x\|=1 \} = \{-1,1\}$$

Sebastien Palcoux

Posted 2010-05-04T21:02:58.510

Reputation: 7 117

2Exact. Moreover, if it were connected, its suspension $\mathbb S^1$ would be simply connected. – ACL – 2016-04-21T06:23:31.800


As a student, I thought (for quite a while) that our textbook had stated that tensoring commutes with taking homology groups. It wasn't until calculating the homology groups of the real projective plane over rings Z and Z/2Z that I realized my mistake.

Herman Jurjus

Posted 2010-05-04T21:02:58.510

Reputation: 168


Two very common errors I see in (bad) statistics textbooks are

(i) zero 3rd moment implies symmetry (though generally stated in terms of "skewness", where skewness has just been defined as a scaled third moment)

(ii) the median lies between the mean and the mode

(I have seen a bunch of related errors as well.)

Another one I often see is some form of claim that the t-statistic goes to the t-distribution (with the usual degrees of freedom) in large samples from non-normal distributions.

Even if we take as given that the samples are drawn under conditions where the central limit theorem holds, this is not the case. I have even seen (flawed) informal arguments given for it.

What does happen is (given some form of the CLT applies) Slutzky's theorem implies that the t-statistic goes to a standard normal as the sample size goes to infinity, and of course the t-distribution also goes to the same thing in the limit - but so, for example, would a t-distribution with only half the degrees of freedom - and countless other things would as well.

The first two errors are readily demonstrated to be false by simple counterexample, and to convince people that they don't have the third usually only requires pointing out that the numerator and denominator of the t-statistic won't be independent if the distribution is non-normal, or any of several other issues, and they usually realize quite quickly that you can't just hand-wave this folk-theorem into existence.


Posted 2010-05-04T21:02:58.510

Reputation: 1

In the statistics text at the college where I teach, (ii) is universal among the examples given, so I formulated the conjecture; but when I tried to prove it and thought about what the mode really is, I realised how badly behaved that can be and found immediate counterexamples. (Then this gets me wondering why anybody would bother using the mode as a statistic for anything, since it's pretty much meaningless, but never mind.) – Toby Bartels – 2011-04-04T09:24:11.057

Toby: sure, you use the mode for cases when the domain of the measurement is not an ordered set but just a set without structure and so the median wouldn't make sense. – Zsbán Ambrus – 2011-04-07T12:01:11.197


False belief: ${\cal P}(\omega)$ has only countable chains with respect to $\subseteq$.

It seems mind-boggling to me that you can start with $\emptyset$, and "add stuff" uncountably many times until you reach $\mathbb{N}$ itself! I learnt this today in a comment by Andreas Blass that he wrote referring to this question.

Dominic van der Zypen

Posted 2010-05-04T21:02:58.510

Reputation: 10 887

2Well, if "adding stuff ... many times" is meant in an iterative sense, i.e., is parametrized by an ordinal, then your original intuition would be right. – Todd Trimble – 2017-10-08T02:34:36.350

2On a similar line: the Hilbert space $\ell_2$ has a family of closed linear subspaces, parametrized on the unit interval, and increasing by inclusion. This looks a bit paradoxical. Should we add orthogonal vectors uncountably many times as $t$ increases from $0$ to $1$? No: just think to $L^2[0,t]$ as subspaces of $L^2[0,1]$ for $0\le t\le 1$ – Pietro Majer – 2017-10-22T14:21:31.927


Before reading about it, I really thought that if $f \colon [0,1] \times [0,1] \to [0,1]$ is a function with the following properties:

  1. for any $x \in [0,1]$ the function $f_x\colon [0,1] \to [0,1]$ defined by $f_x(y)=f(x,y)$ is Lebesgue measurable, and also the function $f^y \colon [0,1]\to[0,1]$ defined by $f^y(x)=f(x,y)$ is Lebesgue measurable, for all $y \in [0,1]$;
  2. both $\varphi(x)=\int_0^1 f_x d\mu$ and $\psi(y)=\int_0^1 f_y d\mu$ are Lebesgue measurable.

Then the two iterated integrals $$ \int_0^1\varphi(x)dx \mbox{ and } \int_0^1\psi(y)dy $$ should be equal. This is false (see Rudin's "Real and Complex Analysis", pag. 167), at least if you assume the continuum hypothesis.


Posted 2010-05-04T21:02:58.510

Reputation: 1 955

1For others reading, the hypothesis left off here is that one must assume $f$ is measurable with respect to the product $\mathcal{B}[0,1] \times \mathcal{B}[0,1]$. – nullUser – 2013-07-08T15:39:41.700

2I really like this example from Rudin's book. Do you know if there exist such an example that does not use the continuum hypothesis (or if it's even possible to find one)? – Kalim – 2010-07-28T13:39:17.873

3I don't know, but this could be a good questions for MO! – Ricky – 2010-07-28T14:28:38.380


If $E$ is a contractible space on which the (Edit: topological) group $G$ acts freely, then $E/G$ is a classifying space for $G$.

A better, but still false, version:

If $E$ is a free, contractible $G$-space and the quotient map $E\to E/G$ admits local slices, then $E/G$ is a classifying space for $G$.

(Here "admits local slices" means that there's a covering of $E/G$ by open sets $U_i$ such that there exist continuous sections $U_i \to E$ of the quotient map.)

The simplest counterexample is: let $G^i$ denote $G$ with the indiscrete topology (Edit: and assume $G$ itself is not indiscrete). Then G acts on $G^i$ by translation and $G^i$ is contractible (for the same reason: any map into an indiscrete space is continuous). Since $G^i/G$ is a point, there's a (global) section, but it cannot be a classifying space for $G$ (unless $G=\{1\}$). The way to correct things is to require that the translation map $E\times_{E/G} E \to G$, sending a pair $(e_1, e_2)$ to the unique $g\in G$ satisfying $ge_1 = e_2$, is actually continuous.

Of course the heart of the matter here is the corresponding false belief(s) regarding when the quotient map by a group action is a principal bundle.

Dan Ramras

Posted 2010-05-04T21:02:58.510

Reputation: 5 258

I'm a little confused.

How does requiring that $(e_1, e_2) \mapsto g$ be continuous fix things? In the indiscrete case, this map is continuous (since every map to the group is).

And why isn't $G^i \to G^i/G$ a principal $G^i$--bundle? – Autumn Kent – 2011-03-06T17:52:03.690

The group in this example starts out with some topology. (I guess I didn't specify that I was thinking of a topological group.) If G started with the indiscrete topology, then your commment makes sense, and we would have a principal bundle for this indiscrete group. But if G is not indiscrete, then the map $(e_1, e_2) \mapsto g$ is not continuous as a map into the topological group G. The proof that continuity of the translation map forces this to be a principal bundle can be found in Husemoller's book on fiber bundles (it's not hard). Let me know if this didn't answer your questions. – Dan Ramras – 2011-03-06T19:57:10.133

Oh! You're saying that a point is not a classifying space for G with some other topology. I thought you were saying that $G^i/G$ wasn't $BG^i$. Thanks for the clarification! – Autumn Kent – 2011-03-06T20:01:32.170

Yes, precisely. It's an odd little example, but helpful when people forget to include the proper conditions... – Dan Ramras – 2011-03-06T21:06:13.230

For sure. Thanks again. +1 – Autumn Kent – 2011-03-06T21:10:32.010

1Maybe even more amazing wrong belief in this field: $\dim(E/G)\le\dim E$ (there are counterexamples by A.N. Kolmogorov) – mikhail skopenkov – 2011-06-09T14:52:23.317


Inversion is an automorphism of a group. ('Cause it, like, preserves the conjugacy classes and all that...)

Robert Haraway

Posted 2010-05-04T21:02:58.510

Reputation: 696


Duality reverses inclusions of vector spaces.


Posted 2010-05-04T21:02:58.510

Reputation: 1

9That's funny, because I don't imagine this kind of idea would occur to someone who has just learned the definition of a dual space. That would be a strangely sophisticated mistake to make. – Thierry Zell – 2011-04-07T00:21:04.070

And, once you learn that this is wrong, you can make the opposite mistake. See my comments here http://sbseminar.wordpress.com/2011/02/22/sobolev-spaces-on-manifolds/ on how surprised I was that duality DOES reverse the inclusions between Soboloev spaces.

– David E Speyer – 2011-04-11T12:07:00.350

The mistake is somehow in the wording. The dual of the inclusion morphism is reversed, it's just not an inclusion anymore. – Manuel Bärenz – 2015-09-04T15:42:20.950


A possible false belief is that "a maximal Abelian subgroup of a compact connected Lie group is a maximal torus". Think of the $\mathbf Z_2\times\mathbf Z_2$-subgroup of $SO(3)$ given by diagonal matrices with $\pm1$ entries.

Claudio Gorodski

Posted 2010-05-04T21:02:58.510

Reputation: 3 272

2Fu... I just "proved" that again as an exercise a few days ago. – Johannes Hahn – 2013-03-06T00:02:09.810


I don't know how common this mistake is, but I think it's worth mentioning. I used to think that existence of non-measurable sets is guaranteed by the axiom of choice only.

In the presence of AC, there cannot be a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ that extends the usual Lebesgue measure.

It is true that we cannot extend the Lebesgue measure in a translation-invariant way by various Vitali set constructions. On the other hand, if you do not insist that the extension is translation-invariant, it might be possible to do this relative to a real-valued measurable cardinal assumption.

Theorem (Ulam): If there exists a cardinal $\kappa$ such that there exists an atomless $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$, then there exists a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ extending the Lebesgue measure.


Posted 2010-05-04T21:02:58.510

Reputation: 1 683

I think you need $\kappa\leq\frak c$, no? – Asaf Karagila – 2015-01-22T14:44:37.597

@AsafKaragila: I believe the assumption that our measure is atomless already implies that $\kappa \leq 2^{\omega}$. – Burak – 2015-01-22T14:45:55.950

Take any measurable cardinal, then there is an atomless probability measure on its power set. It's just that an event is either improbable or its negation is improbable. Unless by probability measure you mean it obtains many values, not just two. – Asaf Karagila – 2015-01-22T14:48:22.503

Isn't that measure atomic if you are deriving it from the ultrafilter? (By an atom, I mean any $A$ of positive measure such that for any $B \subseteq A$ either $\mu(B)=0$ or $\mu(B)=\mu(A)$). I will have to catch a course now but the theorem I referred to should be in Kanomori (indeed, I checked the pdf and it is Theorem 2.5) – Burak – 2015-01-22T14:53:44.407

Ohhhh, right. I was thinking about atoms in the sense of Boolean algebra, as minimal positive elements. Thanks for the clarification! – Asaf Karagila – 2015-01-22T14:55:39.673


Many students believe that every abelian subgroup is a normal subgroup.

Denis Serre

Posted 2010-05-04T21:02:58.510

Reputation: 26 530


A Banach space $X$ is reflexive if it is isomorphic to its double dual ${X^*}^*$.

(Couldn't find this is the list…)


Posted 2010-05-04T21:02:58.510

Reputation: 6 399

3Even isometric fails. (Lindenstrauss & Tzafriri, in the '60s I believe.) – Hachino – 2015-05-12T08:19:04.300


$\mathbb{R}^2$ has a unique complex manifold structure; it's just $\mathbb{C}$ right?

Michael Albanese

Posted 2010-05-04T21:02:58.510

Reputation: 5 261

4Some evidence. – Michael Albanese – 2015-12-24T10:21:38.027


Here's one I was reminded recently during lunch in the common room.

A maximal abelian subalgebra of a semisimple Lie algebra is a Cartan subalgebra.

This is true for compact real forms of semisimple Lie algebras, but fails in general. The missing condition is that the subalgebra should equal its normaliser.

José Figueroa-O'Farrill

Posted 2010-05-04T21:02:58.510

Reputation: 24 255

2``The missing condition is that the subalgebra should equal its normaliser'', or that the subalgebra consists of semisimple elements, no? (That provides another perspective on why it's true for compact real forms.) – LSpice – 2013-12-12T23:26:50.677

For example, the usual proof of Schur's theorem on commutative algebras of matrices produces abelian Lie subalgebras of $\mathfrak{sl}_n$ much larger than the rank of $\mathfrak{sl}_n$. – Mariano Suárez-Álvarez – 2010-08-04T23:41:50.787

Yes, my favourite example (being a physicist) is the stabiliser in $\mathfrak{so}(1,n)$ of a nonzero zero-norm vector in $\mathbb{R}^{1,n}$ for $n>4$. The stabiliser contains an abelian ideal (infinitesimal null rotations) of dimension $n-1$. – José Figueroa-O'Farrill – 2010-08-05T00:02:54.817

I meant to write $n>3$ above. – José Figueroa-O'Farrill – 2010-08-05T00:03:20.010


Consider the following well-known result: Let $(E,\leq)$ be an ordered set. Then the following are equivalent: (i) Every nonempty subset of $E$ has a maximal element. (ii) Every increasing sequence in $E$ is stationary.

It is immediate that (i) implies (ii). To prove the converse, one assumes that (i) is false and then "constructs step by step" a strictly increasing sequence.

The common mistake (which I have seen in textbooks) is to describe the latter construction as a proof by induction. In fact, the construction uses the axiom of choice (or at least the dependent choice axiom).

(As a special case, I don't think ZF can prove that every PID is a UFD.)

Laurent Moret-Bailly

Posted 2010-05-04T21:02:58.510

Reputation: 12 501

It’s not exactly wrong to call it a proof by induction. In ZFC, the proof of dependent choice — or of just about any instance of it, eg the one here — works by combining induction and choice. So I’d agree it’s wrong to sweep the choice under the carpet; but if you’re not explicitly invoking DC, then you will be using induction as well. – Peter LeFanu Lumsdaine – 2010-12-01T15:34:52.210

Peter, let's state DC as follows: "If $(pn:X{n+1}\to X_n)$ is an $\mathbb{N}$-projective system of nonempty sets with all $p_n$ surjective , then projlim($X_n$) is nonempty."

Proof from AC: put $X:=\coprod_{n\geq0}Xn$ and $X^+=\coprod{n>0}X_n$ with obvious map $p:X^+\to X$. Then $p$ is onto, so has a section $s$ (family of sections of all $p_n$'s). Given $x_0\in X_0$, sequence $(s^n(x_0))$ is an element of projlim($X_n$).

I agree that we do need induction to define $s^n$. But iteration of a map is such a basic tool that I don't agree to call any proof using it a "proof by induction". – Laurent Moret-Bailly – 2010-12-07T11:49:07.763


Draw the graph of a continuous function $f$ (from $\mathbb{R}$ to $\mathbb{R}$). Now draw two dashed curves: one which everywhere a distance $\epsilon$ above the graph of $f$ and one which is everywhere a distance $\epsilon$ below the graph of $f$. Then the open $\epsilon$-ball around $f$ (with respect to the uniform norm) is all functions which fit strictly between the two dashed curves.


Posted 2010-05-04T21:02:58.510

Reputation: 193

Hmm, very nice (once clarified to the open ball)! Easily dispelled as soon as you question it, but I could easily imagine using it without thinking and missing the alternation of quantifiers that’s going on under the surface. – Peter LeFanu Lumsdaine – 2010-12-01T15:30:05.853

6Surely this is true if you are talking about the closed ball, and only just barely false for the open ball (and if we were talking about functions from $[a,b]$ to $\mathbb{R}$ it would be true)? Or else I am one of those with the false belief... – Nate Eldredge – 2010-10-10T18:26:43.690

You are right, I should have specified open ball, thanks. I think it is just barely false for the open ball. Honestly, I held this false belief until a couple of days ago, and I haven't thought much about correcting my belief. Probably the real open epsilon ball is the union of all functions that fit between dashed curves a distance strictly less than epsilon away from f? At any rate, I think the above picture is the right way to think about it most of the time. But it gives results such as $tan^{-1}$ being in the open ball of radious pi/2 centered at 0 if you interpret it literally. – user4977 – 2010-10-10T19:24:04.080


This is a common error made by mature mathematicians in many books and papers in analysis, especially in differential equations: If $X$ is a closed subspace of a Banach space $Y$, then the $Y^*$ (the dual of $Y$) is isomorphic to a subspace of $X^*$ (the dual of $X$). It is false (of course) since Euclidian space $\mathbb R$ is a subspace of $\mathbb R^2$, yet the dual of $\mathbb R^2=\mathbb R^2$ is not isomorphic to a subspace of the dual of $\mathbb R=\mathbb R$. I guess, sometimes they really, really want it to be true. Cheers Boris

Boris Shekhtman

Posted 2010-05-04T21:02:58.510

Reputation: 1

I'll take your word for it, but since that statement is false even without introducing norms and topology, it staggers me that people could even believe that. They might say it without thinking, I guess – Yemon Choi – 2010-10-20T02:58:42.067

4I would also be shocked if this really gets believed often! It seems to be a sort of “mis-dualisation”: they dualise “$X$ is a subobject of $Y$” to “$Y^$ is a subobject of $X^$”, where the correct dual is “$X^$ is a quotient of $Y^$”. – Peter LeFanu Lumsdaine – 2010-12-01T15:19:42.620


"If a field $K$ has characteristic 0 and $G$ is a group, then all $KG$-modules are completely reducible."

True for finite groups but very false in general.

Alastair Litterick

Posted 2010-05-04T21:02:58.510

Reputation: 216


1- A very common mistake that 1st year students (but not even a single mathematician) think that it is true is "a transitive and symmetric relation on a set is reflexive". But as the empty set is a transitive and symmetric relation but not reflexive on any non-empty set. Of course there lots of non-trivial examples also.

2- Another common mistake is that the expression "countable union of countable sets is again countable" is independent of axiom of choice (AC). Many people make the proof of this statement without mentioning axiom of choice. Indeed, in his holly book Algebra, Lang proves this statement just by taking an ordering from each countable set and continues without the mentioning AC.


Posted 2010-05-04T21:02:58.510

Reputation: 117

2For big-list questions, it's usually best to post independent answers as separate answers. – Nate Eldredge – 2010-12-02T15:13:53.747

3+1 for #2. Baby Rudin is another offender. And many authors use so-called "diagonalization tricks" for proving compactness theorems like Arzela-Ascoli and Prohorov, which typically reduce to the compactness of $[0,1]^\mathbb{N}$. – Nate Eldredge – 2010-12-02T15:21:36.703

Isn't the more right statement that a transitive and irreflexive relation is assymetric? – Zsbán Ambrus – 2010-12-04T22:46:02.557


I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.


Posted 2010-05-04T21:02:58.510

Reputation: 814

1is it true though if the image is simply-connected? – KotelKanim – 2011-04-17T11:33:58.513

thanks. that's great. I never thought about it before, but it just sounded right... – KotelKanim – 2011-04-19T07:46:53.067

3No true. Take $f(z)=z^3-3z$ and restrict it to the complement of $\lbrace 1,-1\rbrace$ so that $f'(z)$ is never $0$. It maps this domain onto $\mathbb C$. – Tom Goodwillie – 2011-05-04T00:16:32.560

@TomGoodwillie: what if both domain and image are simply-connected? – Michael – 2013-12-03T00:45:17.693


This is more of a false philosophy than a clear mistake, but nevertheless it is very common:

A compact topological space must be "small" in some sense: it should be second countable or separable or have cardinality $ \le 2^{\aleph_0}$, etc.

This is all true for compact metric spaces, but in the general case, Tychonoff's theorem gives plenty of examples of compact spaces which are "huge" in the above sense.


Posted 2010-05-04T21:02:58.510

Reputation: 2 414

More or less along the same lines, one is usually lead to think that "every topological space is Hausdorff". – Marco Golla – 2011-05-04T14:43:46.310


Let $B(r_1) \subset B(r_2)$ be two open balls of radius $r_1$ and $r_2$ respectively. Then $r_1 \leq r_2$.

Bounded metric spaces give trivial counterexamples. Also, $B \left( \frac{1}{6}, \frac{2}{3} \right) \subsetneq B \left( \frac{1}{2}, \frac{1}{2} \right)$ in $(0,+ \infty)$.


Posted 2010-05-04T21:02:58.510

Reputation: 1 337

What is $B(a,b)$ here? – Turbo – 2013-12-03T10:09:04.973

Here, $B(a,b)$ is the open ball of radius $b$ centered at $a$. – Seirios – 2013-12-03T21:41:35.993


"Let $E$ be a complete locally convex topological vector space (or a complete topological vector space or a complete topological group) and let $F$ be closed vector subspace (or a closed subgroup). Then the quotient $E/F$ is complete."

This just has to be true. One can almost see the proof. And in fact it is true for Banach spaces. So it has to be true for locally convex spaces as well.

Another one with completions:

"Every topological group is a dense subgroup of a complete topological group." True for abelian groups but false in general (take the homeomorphism group of $[0,1]$ with the compact open topology)


Posted 2010-05-04T21:02:58.510

Reputation: 487


When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.

I took some time to discover that my mistake was that I thought about $\mathbb R$ over $\mathbb Q$ as a Banach space. In fact, this vector space is a complete metric space (in the sense of Cauchy sequences), but I realized later that the word Banach space is reserved only for vector spaces defined over the fields $\mathbb R$ or $\mathbb C$.


Posted 2010-05-04T21:02:58.510

Reputation: 1 242

10You can define Banach spaces over any complete field. For example, one can define p-adic Banach spaces. But Q isn't complete with respect to any of its norms. – Qiaochu Yuan – 2010-05-04T22:14:58.850

2In fact the Theorem I mentioned in my answer, is based on the Riesz lemma and this lemma is not valid if the scalar fields is not complete. – Leandro – 2010-05-04T22:34:38.510

1@QiaochuYuan ...unless you use the trivial norm. – Mario Carneiro – 2015-10-20T21:16:06.893


Here's a mistake I've seen from students taking a first course in linear analysis. For a vector $g$ in a Hilbert space $H$, it is true that $\langle f,g\rangle=0$ for every $f\in H$ implies $g=0$. This leads us to the mistaken:

“Let $(g_n)$ be a sequence in $H$. If, for every $f\in H$, $\langle f,g_n\rangle\to0$, then $g_n\to 0$.”


Posted 2010-05-04T21:02:58.510

Reputation: 986

You wrote:

"Here's a mistake I've seen from students taking a first course in linear analysis."

Then you wrote:

"For a vector $g$ in a Hilbert space, $\langle f,g\rangle$ for every $f \in H$ implies $g = 0$."

At this point the reader could be wondering what that is a mistake. – Michael Hardy – 2010-12-01T22:35:55.660

....sorry; I meant "$\langle f,g \rangle = 0$ for every[....]" – Michael Hardy – 2010-12-01T22:36:37.470

2@Michael: all answers are CW; so if we think some wording needs clarifying, we can do it ourselves! – Peter LeFanu Lumsdaine – 2010-12-02T00:43:12.583


I saw many students using the "fact" that for a subset $S$ of a group one has $SS^{-1}=\{e\}$


Posted 2010-05-04T21:02:58.510

Reputation: 101

5This is an interesting example, because it addresses the mistakes that come from the all-too frequent confusion with notations. But we need our shortcuts, our $f^{-1}(x)$ versus $x^{-1}$, etc. Obtaining concise notations while avoiding confusion: a tricky proposition! – Thierry Zell – 2011-04-14T15:50:49.610


If $\alpha>0$ is not an integer, the set of functions $f:[a,b]\rightarrow\mathbb R$ such that $$\sup_{y\ne x}\frac{|f(y)-f(x)|}{|y-x|^\alpha}<+\infty$$ is ${\mathcal C}^\alpha([a,b])$.

False for $\alpha>1$, because this set contains only constant functions.

Denis Serre

Posted 2010-05-04T21:02:58.510

Reputation: 26 530


While this example was already mentioned by Skopenkov in a comment, here it is, explicitly:

If $G\times X\to X$ is a free continuous action of a compact metrizable group on a compact metrizable space $X$, then $X\to X/G$ is a principal $G$-bundle.

This is true if $G$ is a Lie group (Gleason) and, more generally, for proper Lie group actions on locally compact spaces (Palais), but false in general (the example is essentially due to Kolmogorov), see:

R. F. Williams, A useful functor and three famous examples in topology. Trans. Amer. Math. Soc. 106 (1963) 319–329.

In this example, $\dim(X/G)=2, \dim(X)=1$.

Moishe Cohen

Posted 2010-05-04T21:02:58.510

Reputation: 677


The following seems not to be here yet.


$R[[x_1,x_2,x_3,\dotsc]]/(x_2,x_3,\dotsc)$ isomorphic to $R[[x_1]]$ ${}\hspace{118pt}$ (f)

Source of the misconception. A fallacy of type false generalization: for any $n\in\mathbb{N}$ it is true that

$R[[x_1,x_2,x_3,\dotsc,x_n]]/(x_2,x_3,\dotsc,x_n)\cong R[[x_1]]$ ${}\hspace{125pt}$ (t)

but to conclude from this that (f) was true by passing to the limit $n\to\infty$ is fallacious.

Reason for why the misconception is false. E.g. the formal power series $f:=x_2+x_3+\dotsm$ is an element of $R[[x_1,x_2,x_3,...]]$, but by the standard definition of $I:=(x_2,x_3,\dotsc)$, which after all means nothing more than the $R[[x_1,x_2,x_3,\dotsc]]$-module generated by the infinite set $\{x_i\colon i\in \omega,\ i\geq 2\}$, the ideal $I$ does not contain $f$. (Having coefficients from the huge power series ring $R[[x_1,x_2,x_3,\dotsc]]$ does not help.)

Reason for including the example. I saw this misconception in a dissertation. For obvious reasons, I won't give the source.

Further remarks. In the above, $R$ can be any commutative unital ring, and $R[[x_1,x_2,x_3,\dotsc]]$ as usual means the projective limit in the category of commutative unital rings of the diagram $\dotsm\twoheadrightarrow R[[x_1,x_2,x_3]]\twoheadrightarrow R[[x_1,x_2]]\twoheadrightarrow R[[x_1]]$ consisting of the canonical projections.

Peter Heinig

Posted 2010-05-04T21:02:58.510

Reputation: 3 909


Multiplication of differential forms is inherently anti-commutative. Thus, if $x$ and $y$ are coordinates on a surface, then $dx \wedge dy$ makes sense but $(dx)^2+(dy)^2$ is either nonsense or, if it means anything, is $0$.

I'm not sure why I believed this, but I did for several years. I tried my best to avoid creating this impression in my students, but I think it still happened in some of them, simply because the curriculum spends a lot of time on integration and Stokes theorem and very little time on metrics, curvature, etc.

David E Speyer

Posted 2010-05-04T21:02:58.510

Reputation: 98 510


I have checked the existing answers, but I think this one is not given yet. Sorry, if I missed it.

Although the incompleteness theory of Gödel is generally correctly understood, the consequence of it has multiple false beliefs:

  • Due to the incompleteness theory it is not possible to make an AI. Humans will always be be superior to the AI. This assumes that human thinking is complete and will eventually find the answer on any question.

  • Due to the incompleteness theory, it is not possible to formalize mathematics. This is refuted by many proof systems, which can formalize almost all mathematics.

As side note, I think this is partly fueled how logic is taught. It puts more emphasis on impossibilities (incompleteness theory), than possibilities (a proof system).

Lucas K.

Posted 2010-05-04T21:02:58.510

Reputation: 1 023

+1. I always found it wrong that classes in logic put so much emphasis on negative results. (And I wish they had prepared me better for proof assistants... though I guess one semester does not suffice for the ones that exist today.) – darij grinberg – 2015-09-05T22:17:55.283

There's arguably too much fascination with incompleteness and not enough with completeness, which is more of a cornerstone of model theory. – Todd Trimble – 2015-09-06T01:50:10.340


Every matrix is the sum of a symmetric and an antisymmetric matrix. Hence:

If $V$ is a vector space and $k$ is a number, then the $k$-th tensor product of $V$ with itself decomposes as a direct sum into symmetric and antisymmetric tensors: $$ \underbrace{V \otimes ... \otimes V}_{k\text{ times}} = \Lambda^kV \oplus \mathrm{Sym}^kV $$

Recall (in the finite-dimensional case) the dimensions: $$ \dim \Lambda^k V = \binom{n}{k} \quad\text{ and }\quad \dim\mathrm{Sym}^kV = \binom{n+k-1}{k} $$

Looking at $k=1$ shows that we have non-trivial intersection.

Looking at $n=k=3$ shows that the sum is not exhausting.

Konrad Waldorf

Posted 2010-05-04T21:02:58.510

Reputation: 2 642

2Is this a common false belief? – Jim Conant – 2015-10-18T02:52:24.240

@JimConant: I believed this once. Of course, if you count dimensions it's obviously false. But if $V$ is an infinite-dimensional Hilbert space it sure seems natural to decompose the full tensor product into its Bosonic and Fermionic parts, and you might not think right away to ask whether it works in finite dimensions. That's my excuse, anyway! – Nik Weaver – 2016-03-04T04:22:55.620


A polynomial $p(x)$ of degree $n$, with coefficients in a commutative ring $R$, has at most $n$ roots, counting multiplicity. This is true if $R$ is an integral domain, but it can fail in the presence of zero divisors.

For instance, $p(x) = x^2+5x$ has four solutions when $R = \mathbb{Z}/6\mathbb{Z}$. I realized this mistake when a colleague asked me about factorization over a non-commutative ring, and I realized that I did not even know what would happen in the presence of zero-divisors.

This does motivate a question I have not found an answer to: is the number of solutions of $p(x)$ bounded by a function of the degree and the characteristic of the ring?

Jacob White

Posted 2010-05-04T21:02:58.510

Reputation: 201

6To answer your last question, $x^2-1$ has an infinite number of roots in $k^{\mathbb{N}}$ for any nonzero ring $k$. So, no. – Gro-Tsen – 2016-04-20T17:00:59.567

2Doesn't it only have one solution in $k^\mathbb{N}$ if $k$ has characteristic 2, @Gro-Tsen? – Omar Antolín-Camarena – 2016-04-20T17:36:35.127

2@OmarAntolín-Camarena Oh right, what I wanted to write was $x^2-x$, and I got confused between "idempotent" and "one-potent"(?). But of course $x^2-1$ also works provided, as you point out, that $1\neq -1$ in $k$. – Gro-Tsen – 2016-04-20T20:31:07.867


Taylor's Formula and displacement operator: I (too often) see in papers (mathematical physics but a recent paper (a) by mathematicians also) the statement

False belief 1 : a) Let $D=\frac{d}{dx}$ be the derivation operator. Then, for all $f\in C^\infty(\mathbb{R})$, $$ e^{tD}[f](x)=f(x+t) $$

which is false (take any $\phi\in C^\infty(\mathbb{R})$ with compact support, for instance).

False belief 2 : b) In the same vein, for formal power series (``our object is formal then we do not have to ensure convergences''). Let $S(x)\in \mathbb{R}[[x]]$ ($x$ is a formal variable) then for $t\in \mathbb{R}$, one has $$ e^{tD}[S](x)=S(x+t) $$

which is false as we must have $t$ in the domain of convergence of $S$.

Remarks (i) The function $f\in C^\infty(\mathbb{R})$ is analytic over $\mathbb{R}$ iff $$ (\forall x\in \mathbb{R})(\exists R>0)(\forall t\in ]-R,R[) (\sum_{n\geq 0}\frac{t^n}{n!}D^n[f](x)=f(x+t))\qquad (1) $$ (ii) Even if $f\in C^\omega(\mathbb{R})$, it can happen that the left hand side of eq. (1) do not converge otherwise $f$ would be the restriction of an entire function (which e.g. $\frac{1}{1+x^2}$ is not, for example).

(iii) Even if the LHS of (1) converges for all $x,t\in \mathbb{R}$, $f$ need not be analytic. Consider the following function (classic in theory of distributions)
$$ f(x)=0\mbox{ if } x\notin ]-1,1[\mbox{ and } f(x)=e^{\frac{1}{1-x^2}} \mbox{ if } x\in ]-1,1[ $$ (iv) In the (b) case $S=\sum_{n\geq 0}n!\, x^n$ for example cannot be displaced.

Duchamp Gérard H. E.

Posted 2010-05-04T21:02:58.510

Reputation: 2 035

What is the correct statement? – ಠ_ಠ – 2017-07-13T07:33:31.300

@ಠ_ಠ: it is true for real analytic functions defined on the entire real number line. That is one possible choice of correct statement. – Ben McKay – 2017-07-13T08:07:28.413

1Maybe you could point out the error in my reasoning: for a Lie group $G$, $G$ has a canonical left action on $C^\infty(G, \mathbb{R})$ by $(g.f)(x) =f(g^{-1}.x)$. Since $D=\frac{d}{dx}$ is a left-invariant vector field on $(\mathbb{R}, +)$, then $(e^{tD}.f) (x)=f(e^{-tD}.x)=f(x-t)$. – ಠ_ಠ – 2017-07-13T08:35:50.497

@ಠ_ಠ I added it. The fact that the LHS converges for some $t$ means that $f$ coincides with the sum of its Taylor series in the open disk of center $x$ and radius $|t|$. – Duchamp Gérard H. E. – 2017-07-13T19:52:00.707

I think we probably mean different things with our notation. For me, $\exp: \mathfrak{g} \to G$ always denotes the exponential map of the Lie group, which always exists. But it looks to me like you mean by this notation that you are integrating the Lie algebra action rather than the Lie algebra. – ಠ_ಠ – 2017-07-13T21:02:45.203

@ಠ_ಠ You wanted me to locate the exact "point of belief" in your reasoning, it is when you note $(e^{tD}.f)(x)$ at the beginning of the last chain of equalities. This term does not always exist (only if $f$ is analytic). – Duchamp Gérard H. E. – 2017-07-14T02:00:43.707

2@ಠ_ಠ I think I see what you meant. In the (Fréchet) space $C^\infty(\mathbb{R})$, the evolution equation $$ y(0)=f ; y′(t)=D[y(t)] $$ has a (unique) solution $y(t)[x]=f(x+t)$ which we could (legitimately ?) note $y(t)=e^{tD}[f]$, but one must keep in mind that, in this case $e^{tD}$ cannot be developed without caution. – Duchamp Gérard H. E. – 2017-07-16T14:17:07.437


False belief. If a family $(x_n)_{n\geq 1}$ is commutatively convergent (i.e. summable) in a normed space $(V,\|\ \|)$ then $$ \sum_{n\geq 1} \|x_n\|<+\infty\ . $$

This is true in finite dimensions and has counterexamples in infinite dimensions. Details and counterexamples can be found there.

Recall that a family $(x_i)_{i\in I}$ is called summable with sum $S$ iff $$ (\forall \epsilon>0)(\exists F\subset_{finite} I)(\forall F_1\subset_{finite} I)(F\subset F_1\Longrightarrow \|S-\sum_{i\in F_1}x_i\|<\epsilon) $$ This is equivalent with commutative convergence in case $I\subset \mathbb{N}$ is infinite.

Duchamp Gérard H. E.

Posted 2010-05-04T21:02:58.510

Reputation: 2 035

2I particularly like Grothendieck's version of this. He proves that unconditional convergence is the same as absolute convergence for a locally convex space if and only if it is nuclear, i.e. every continuous linear map to a normed space is a nuclear map. The falsity of this belief then follows from the fact that a nuclear normed space is finite dimensional. – Robert Furber – 2018-02-17T07:44:43.817

@RobertFurber Thank you for this learned description (+1) – Duchamp Gérard H. E. – 2018-02-17T08:03:23.960


An elementary false belief in elementary number theory: for $a, b, c\hspace{.1cm}\varepsilon\hspace{.1cm} \mathbb{N}$

$LCM\left(a,b\right)\times GCF\left(a,b\right) = ab$ .

Thus, $LCM\left(a,b,c\right)\times GCF\left(a,b,c\right) = abc$.

In general, $\left(a_1,a_2,\ldots,a_n\right)[a_1,a_2,\ldots,a_n] = a_1a_2\ldots a_n$.


Posted 2010-05-04T21:02:58.510

Reputation: 1 146

Does GCF mean gcd (greatest common divisor) here? – Zsbán Ambrus – 2010-11-27T19:33:21.010

Yes. GCF means the same: Greatest Common Factor. – Unknown – 2010-11-29T06:46:08.150

17This kind of stuff is easy to rule out, though; it's dimensionally inconsistent. Replacing a, b, c by ka, kb, kc leads to a quick contradiction. – Qiaochu Yuan – 2011-02-24T21:08:05.713

1@Qiaochu, that is a nice quick check. Let the RCF(remnant common factor) be the leftover factor that would make the above second equality true. There seems to be no interesting way of determining $RCF\left(a,b,c\right)$ so that $LCM\left(a,b,c\right)\times RCF\left(a,b,c\right) \times GCF\left(a,b,c\right) = abc$ – Unknown – 2011-02-24T22:05:28.303

2@Elohemahab: actually, the correct generalization is $\gcd(a, b, c) \text{lcm}(ab, bc, ca) = \text{lcm}(a, b, c) \gcd(ab, bc, ca) = abc$. – Qiaochu Yuan – 2011-05-08T23:47:35.220

@Qiaochu, glad to see that. We do have more concerning generalisations to three variables: $$gcd(lcm(a,b),lcm(a,c),lcm(b,c))=lcm(gcd(a,b),gcd(a,c),gcd(b,c))$$

$$gcd(a, lcm(b,c)) = lcm(gcd(a,b),gcd(a,c))$$

$$lcm(a, gcd(b,c)) = gcd(lcm(a,b),lcm(a,c))$$ – Unknown – 2011-05-21T06:24:10.847

though the product is not necessarily $abc$, but it can be made so. – Unknown – 2011-05-21T06:25:19.443

@Qiaochu: I thought the correct generalization of$$\operatorname{lcm}(a,b)=\frac{a\cdot b}{\operatorname{gcd}(a,b)}$$was $$\operatorname{lcm}(a,b,c)=\frac{a\cdot b\cdot c\cdot \operatorname{gcd}(a,b,c)}{\operatorname{gcd}(a,b)\cdot \operatorname{gcd}(a,c)\cdot \operatorname{gcd}(b,c)}$$Another false belief I guess. Maybe not a common one. – bof – 2015-10-18T09:35:23.233


Here's another howler some people commit: If $m$, $n$ are integers such that $m$ divides $n^2$ then $m$ divides $n$.

It's true sometimes, for example if $m$ is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers $p$, $q$, $r$ with $p$ squarefree such that $ m = p q^2 $ and $ n = p q r $

The usual counterexample is that $8$ divides $4^2$ but not $4$ ;-)

John R Ramsden

Posted 2010-05-04T21:02:58.510

Reputation: 836

9An even more trivial counterexample is that 4 divides 2^2 but not 2 :-P – Peter LeFanu Lumsdaine – 2011-02-23T09:40:22.320


For a bounded subset of a metric space the diameter is two times the radius!

Let $S\subset X$ be bounded. The definitions are:

$\mathrm{diameter}(S):=\sup\{d(x,y)\,|\,x,y\in S\}$

$\mathrm{radius}(S):=\inf\{r>0\,|\,\exists x\in X:\,S\subset B(x,r)\}$

where $B(x,r)$ denotes the open ball of radius $r$ around $x$.


Posted 2010-05-04T21:02:58.510

Reputation: 101

4Hםw do you define the radius of an arbitrary bounded subset? – Mark – 2011-04-11T15:34:41.780

1Disproved nicely by Reuleaux triangles. – darij grinberg – 2011-04-12T08:10:18.940

15Disproved nicely by a two-point metric space. – Tom Goodwillie – 2011-04-17T01:36:25.767

An equilateral triangle in the Euclidean plane also does the job (diameter $1$ and radius $1/\sqrt{3}$): $2/\sqrt{3} > 1$. – Jean Van Schaftingen – 2013-11-06T15:07:15.220


Here are some various examples (I hope that some of them weren't already mentioned):
1. If a space $X$ have two different norms $\| \cdot \|_i, i=1,2$ such that $\| \cdot \|_1 \leq \| \cdot \|_2$ then the completion with respect to $\| \cdot \|_1$ is contained in the completion with respect to $\| \cdot \|_2$.
2. If $M_1,M_2$ are isomorphic modules and $N_1,N_2$ are isomorphic submodules then $M_1/N_1$ and $M_2/N_2$ are isomorphic.
3. If $A,B$ are subsets of topological spaces $X,Y$ (resp.) and $A,B$ are homeomorphic then the closures $\overline{A}$ and $\overline{B}$ are also homeomorphic.
4. The standard construction of adjoining unit to the Banach algebra $A$ yields nothing new if $A$ already was unital.
5. The phrase "a function is almost everywhere continuous" means the same as: "the function is almost everywhere equal to the continuous function".
6. Suppose you are trying to prove that some function space $F$ is complete (say that functions are defined on $X$ and real valued): you take a Cauchy sequence $\{f_n\}_n$ and prove that for each point $x \in X$ the sequence $\{f_n(x)\}_n$ is Cauchy. Then form the completeness of $\mathbb{R}$ you obtain a function $f$. The false belief is that it is now enough to show that $f$ belong to $F$.
7. If you have an ascending family $\{A_i\}_i$ then to obtain it's union $\bigcup_{i}A_i$ it is enough to take some countable subfamily
8. A convergent net $\{x_i\}_i$ in a metric space is bounded and the set $\{x_i\}_i \cup \{x\}$ is compact (where $x$ is the limit).
9. If $D$ is an open dense subset of a topological space $X$ then $card \; D= card \; X$


Posted 2010-05-04T21:02:58.510

Reputation: 2 822


As a sequel of this famous answer on $\dim(U+V+W)$, the following inequality is not true $\forall n \ge 4$:
$$ \dim(\sum_{i = 1}^{n} U_i) \le \sum_{r=1}^{n} (-1)^{r+1} \sum_{i_1 < i_2 < \dots < i_r} \dim(\bigcap_{s=1}^{r}U_{i_s}) = \alpha$$
Darij Grinberg has found a counter-example (see this post).

Same flavor: for $n \le 5$, it is true that $\alpha \ge 0$ (see this proof), but it's false for $n>5$ (see this comment).

Sebastien Palcoux

Posted 2010-05-04T21:02:58.510

Reputation: 7 117


Here is a short list of some false beliefs I had when I was studying mathematics, I suppose they may be common but I have never checked:

  • I was in the last year of high school and studying university-level math in advance. I remember trying for a week to prove that a continuous injective map from an open subset of $\mathbb{R}^2$ to $\mathbb{R}^2$ that preserves "being aligned" (I mean that maps aligned triples to aligned triples) must be the restriction of an affine map (over $\mathbb{R}$). That is disproved by restrictions of projective transformations... Which I knew of but I was not able to see they contradicted my belief. When my teacher told me "What about projective transformations?"... I felt dumb.
  • I was in the 1st year of PhD studies. My advisor, Adrien Douady, had an idea to build polynomial Julia sets with positive Lebesgue measure. Julia sets are fractals, often with complicated topological structures at every scale. Surely that must be the source of measure? So as an exercise, I tried for a week to prove that Jordan curves are necessarily of Lebesgue measure 0. I told Adrien about my attempts. He gave me a counter-example. I felt dumb.
  • Learning that there are closed subsets of the interval with positive Lebesgue measure but no interior did not surprise me as much, as the construction is very simple, but still that's a bit counterintuitive.
  • When you zoom on the Mandelbrot set, you see all that round components with smooth boundaries. They look so round. Surely they must be circles, for otherwise the difference would be visible. Well... they are not (except one). Guess how I felt when I learned.
  • Frankly, when learning the first time about complex numbers, did anybody here expect that, adding the square root of -1 to the reals would add the roots of all other polynomials?
  • I was giving a lecture to math teachers about sensitivity to initial condition (call it chaos) and showing strange attractors on the computer, one told me that by the very presence of chaos, what we see may be quite far from the actual behaviour of the equation, save reality. It turns out hyperbolic systems are stable, so I believe this is still representative (it does not prove it but it is an encouraging hint).
  • ... Chaos in deterministic systems. I won't develop on that.
  • Surely before hearing of set theory and Cantor's argument, you will believe that all sets are countable. Then after learning that this is not the case, you will think that $\mathbb{R}^2$ must be bigger than $\mathbb{R}$, right?
  • You have a $C^\infty$ function on the right half plane, all of whose derivatives have a continuous extension to the boundary line. Surely, it must be easy to extend it to a $C^\infty$ function of the whole plane, isn't it? Well... You can but I would not call it easy.
  • Short statements have short proofs. Disproved by Fermat's last theorem (among others).
  • I was quite disappointed to learn that there cannot be a finite non-commutative field (division algebra).

I have a few other examples, that I would not term "common false beliefs" but rather "fun and surprising math facts". Is there already a MO question about that?

Arnaud Chéritat

Posted 2010-05-04T21:02:58.510

Reputation: 920


Probably my fault for not paying enough attention in analysis, but:

Any continuous function on the interval that has derivative equal to zero almost everywhere is constant.

Aaron Bergman

Posted 2010-05-04T21:02:58.510

Reputation: 2 273


For awhile, I used to think:
If $depth\ M\ge depth\ N$ then $depth\ M_p\ge depth\ N_p$; for any prime ideal $p$ and finite R-modules $M$ and $N$ (Which is not true).

user 1

Posted 2010-05-04T21:02:58.510

Reputation: 737


Many people believe that Cantor proved the uncountability of the real line using a diagonal argument. This paper does not that provide that proof; Cantor's stated purpose was to prove the existence of `uncountable infinities' without using the theory of irrational numbers.

KP Hart

Posted 2010-05-04T21:02:58.510

Reputation: 2 939

2The link in the answer goes to the wrong page - it should go to page 75, not page 72. – David Roberts – 2012-06-13T06:41:02.457

1And it looks like a diagonal argument to me. – David Roberts – 2012-06-13T06:43:29.850

2More to the point, I think, is that the paper proves that the power set of any set has greater cardinality than the set itself. This is the first proof that there is no greatest cardinality. (The uncountability of the real line easily follows, even if Cantor does not mention it because he has bigger fish to fry.) – John Stillwell – 2010-05-31T05:12:37.627

2Just to fill in some history here: if I remember right, Cantor first proved the uncountability of the reals by other arguments, then later (as you reference) found the diagonal argument, as a proof of the more general statement about power sets. – Peter LeFanu Lumsdaine – 2010-09-27T03:01:15.367


"The set A = {a, b} has two elements..."

It's quite simple to notice that a can be the same as b, but after 5 years of university there were people still believing it...

Tom Pinkeith

Posted 2010-05-04T21:02:58.510

Reputation: 1

I am confused here, as there is no suggestion that a,b are variables, since there are no quantifiers. It seems to me this is true since the first two letters of our alphabet are indeed distinct. – roy smith – 2010-12-01T19:34:12.477

E.g. if you said something like "for all a,b, (in some given universe) the set {a,b} has two elements", then I would agree. – roy smith – 2010-12-01T19:35:03.380

2Single-letter symbols are usually assumed to be variables, if the context doesn't determine otherwise, even in the absence of quantifiers. (You can put in an implicit universal quantifier to close up all sentences.) – Toby Bartels – 2011-04-04T09:41:17.470

@TobyBartels, I think that this statement can be believed only if one is willing to speak of the 2-variable polynomial $e\pi$. (Perhaps you regard $e$ and $\pi$ as part of a universal context. :-) ) – LSpice – 2013-12-14T22:26:21.903

1In a context where one is discussing real analysis, $e$ and $\pi$ are generally taken to be the famous constants. But this is hardly universal; in other contexts, they may have very different meanings. – Toby Bartels – 2013-12-15T05:16:27.640

3Here is a related but slightly less obvious situation. The ordered pair $(a,b)$ is generally defined in set theory to be ${{a},{a,b}}$. This is generally thought of as a set with two elements. But what if $a=b$? – Bruce Blackadar – 2014-09-26T04:12:20.760

21I'm not sure there is a false belief here, as much as awkward writing. Depending on context, I might very well write "The set ${a,b }$ (where $a$ and $b$ might be equal)..." if this issue mattered. – David E Speyer – 2010-05-06T11:16:34.123

1There are many situations where one needs to speak of a set of two numbers that may or may not be equal. E.g.: "Let x<sub>1</sub>, x<sub>2</sub> &isin; ℝ. Then among all the open intervals containing the set {x<sub>1</sub>, x<sub>2</sub>}, none of them is contained in all the others."

If one is addressing mathematicians, there is no need to specify that x<sub>1</sub> might be equal to x<sub>2</sub>. – Daniel Asimov – 2010-06-17T23:34:28.633


Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.

The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.

This statement happens to be true.

Here is a reference:


The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.

Here is the (trivial) algebra background.

Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.

We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.

Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.

Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.

Pierre-Yves Gaillard

Posted 2010-05-04T21:02:58.510

Reputation: 1 435


True: Given a graded algebra $A$, there is a notion of a "homogeneous" ideal of $A$. It is a property that connects an ideal of $I$ with the grading and is often necessary to require. For example, if $I$ is a homogeneous ideal of $A$, then the algebra $A / I$ is graded again. If $I$ is not homogeneous, then it is not graded in general (since the projections of different graded components of $A$ onto $A / I$ might have nonzero intersection).

False: Given a filtered algebra $A$, there is a notion of a "filtered" ideal of $A$.

There is no such notion. We can require $I$ to be generated by $I\cap A_n$ for some $n$, or actually to lie inside $A_n$ for some $n$, but in most cases none of these is actually needed. (Correct me if I am wrong.) Formulations like "Let $I$ be an ideal compatible with (or respecting) the filtration" are cargo cult.

But: Given a filtered algebra $A$ and a generating set $G$ of an ideal $I$ of $A$, it is an important question whether $I\cap A_n$ is generated by $G\cap A_n$ for every $n\in \mathbb N$. This is not always satisfied, often nontrivial (in many cases it can be proved by using the diamond lemma to show that every element of $A_n$ has a unique "remainder" modulo $I$ in a certain sense, and this remainder can be obtained by repeated subtraction multiples of elements of $G\cap A_n$) and used tacitly in various texts.

darij grinberg

Posted 2010-05-04T21:02:58.510

Reputation: 15 489

Good point, but "cargo cult"? – Tom Goodwillie – 2011-03-15T14:32:51.360

3What I mean is: People use these formulations as a protective charm against a danger they don't see but intuitively feel is there, although closer inspection shows that it is pure superstition. – darij grinberg – 2011-03-15T17:26:25.660


A set is compact iff it is closed and bounded.

Mustafa Said

Posted 2010-05-04T21:02:58.510

Reputation: 973

4This is perhaps a common false belief among undergraduates, but one that is dispelled by just a superficial acquaintance with functional analysis. – Todd Trimble – 2013-12-09T02:45:55.013

@ Todd Trimble: true, but then also the belief about $sin$ suggested by the OP is only common among people who have not completed a course in complex analysis. – Delio Mugnolo – 2013-12-13T08:34:43.317

I thought "bounded" is only defined on metric spaces, and this is true on metric spaces. Is that wrong? – Akiva Weinberger – 2015-09-01T02:48:28.097

I have seen analysis textbooks take this as a definition. I hope they realize that they are contributing to future confusion in their readers once they move on to topology or even metric spaces. @AkivaWeinberger, The Heine-Borel theorem stated in this way makes sense for arbitrary metric spaces, but it is only true for complete metric spaces for which balls are totally bounded. The correct statement of H-B for general metric spaces is "a metric space is compact iff it is complete and totally bounded". – Mario Carneiro – 2015-10-20T21:27:56.063

@AkivaWeinberger: Yes, it is wrong. The closed unit ball of an normed vector space is compact if and only if the space is finite dimensional. – ACL – 2016-04-21T13:43:28.687


Some undergraduate common false beliefs that I found

(1) If $H$ is a subgroup of $\mathbb{Z}$ and $H$ and $\mathbb{Z}$ are isomorphic, then $H = \mathbb{Z}$;

(2) In a metric space every two open balls are homeomorphic;

(3) For $p \in [1, \infty]$, $L^p(X, \mathfrak{M}, \mu) = \left\{ f \in \mathbb{C}^X : \int_X |f|^p \, d \mu < \infty \right\}$ is a $\mathbb{C}$-normed vector space, with the norm $\lVert f \rVert_p = (\int_X |f|^p \, d \mu)^{1/p}$.

Belief (1) is very naive, for every nontrivial subgroup of $\mathbb{Z}$ is of the form $n \mathbb{Z}$, all of them isomorphic with $\mathbb{Z}$. For (2) people tend to think of normed vector spaces and forgets the discrete metric spaces. For (3) some people just forget that one have to consider the quotient space, where the classes $[f]=[g]$ iff $f=g$ $\mu$-almost everywhere.

Belief (1) is very naive, because every nontrivial subgroup of $\mathbb{Z}$ is of the form $n \mathbb{Z}$, all of them isomorphic to $\mathbb{Z}$. For (2) people tend to think of normed vector spaces and they forget the discrete metric spaces. For (3) some people just forget that one have to consider the quotient space, where the classes $[f]=[g]$ iff $f=g$ $\ \mu$-almost everywhere.


Posted 2010-05-04T21:02:58.510

Reputation: 121

Well, in (1) I think I can replace $\mathbb{Z}$ by an arbitrary group $G$, because $\mathbb{Z}$ do not come in mind so quickly. – Gustavo – 2016-01-08T04:03:47.173


Anytime I wanted to write an answer to this question, I doubted maybe it is not as common as worthy of mentioning here. In fact, I am also not sure how common is the false belief that I observed today in a PDE class. I didn't observe that in many years of teaching calculus, but today four or five students in a small PDE class when calculating a definite integral by parts only applied the limits of the integral to the "second" integral, that is:

$$\int_{a}^b{f(x) g'(x) dx}=f(x) g(x) - \int_{a}^b{f'(x) g(x) dx}$$

Haven't I observed well enough in my calculus classes?

Amir Asghari

Posted 2010-05-04T21:02:58.510

Reputation: 1 157

1Definitely integrals are numbers and $f(x)g(x)$ is a function of variable $x$. Formula as written is something very strange. – Fedor Petrov – 2016-04-20T18:58:19.420

1@FedorPetrov More strange is that most students don't see such a very strange something :) – Amir Asghari – 2016-04-20T19:50:24.163


Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references



Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:

There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\\ E_s\\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

Pierre-Yves Gaillard

Posted 2010-05-04T21:02:58.510

Reputation: 1 435

3"$\emptyset$ is a basis for {0}" is an immediate consequence of the definitions. There is no false belief in this point. – Johannes Hahn – 2010-05-12T15:14:54.070

2Dear Johannes, please reread my post. – Pierre-Yves Gaillard – 2010-05-12T15:18:14.740

Even a cursory examination of Nick Higham's book


will show that both these opinions on the evaluation of matrix exponentials are hopelessly naive.

– Robin Chapman – 2010-05-15T09:17:03.297

Dear Robin, Thanks for your answer. I don't have Nick Higham's book. I was wondering if you could be more precise. Your comment is very surprising to me: I thought I was stating a triviality. Here are two references

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

Looking forward to hearing from you.

– Pierre-Yves Gaillard – 2010-05-15T10:51:26.790

Higham has a whole chapter (10) on the matrix exponential, most of which can be found on Google books. – Robin Chapman – 2010-05-15T12:03:23.830

I looked at Chapter 10, but I still don't see your point. I wrote "Let f(z) be the minimal polynomial of A, let g(z) be f(z) times the singular part of e^z/f(z), and observe e^A=g(A)." Are you claiming that Higham disproves this statement? In this case, could you say where? Are you claiming that this statement is false? In this case, could you say why? – Pierre-Yves Gaillard – 2010-05-15T13:04:38.037

5Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. – Robin Chapman – 2010-05-15T14:07:02.857

I agree with your criticism. The point I awkwardly tried to make is this. Computing a matrix exponential and computing the singular part of the quotient of the exponential function by a nonzero polynomial are equivalent problems (in the sense that there is a simple dictionary between them). – Pierre-Yves Gaillard – 2010-05-15T16:09:57.450

1I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. – Qiaochu Yuan – 2010-05-17T06:19:55.397

Dear Qiaochu Yuan: What is the OP? I think the way matrix exponential is usually presented, that is by invoking Jordan decomposition, is based on a conceptual error. This conceptual error (in my opinion) is the failure to see that the exponential of a matrix is obtained by evaluating on this matrix the product of the minimal polynomial f by the singular part of e^z/f(z). This statement is much simpler than the Jordan Decomposition Theorem. – Pierre-Yves Gaillard – 2010-05-17T15:06:46.900

OP = original poster, in this case, gowers. – JBL – 2010-05-19T03:32:56.960

It would be better to post these as separate answers, so that they may be voted and commented on independently. – Nate Eldredge – 2010-05-19T14:50:57.413

Dear Nate, I followed your advice. Thanks! – Pierre-Yves Gaillard – 2010-05-19T16:13:34.567


I'm seven years late to the game, but here is mine:

False belief: The irrational numbers, in their usual topology as a subset of $\mathbb{R}$, are not a complete metric space.

Pace Nielsen

Posted 2010-05-04T21:02:58.510

Reputation: 6 596

3Could you write that more carefully? You mean the false belief to be that the irrationals, as a topological space, can't be complete for some metric. What you wrote can be easily confused with saying the irrational numbers are not complete for the usual metric coming from $\mathbf R$, which is true rather than false. Consider the "false belief" that $(-1,1)$ with its topology from $\mathbf R$ can't be made into a complete metric space for some metric. Certainly it's not complete for the usual metric, but it is if we use $\tan(\pi x/2)$ to identify $(-1,1)$ with $\mathbf R$ topologically. – KConrad – 2017-07-15T03:26:01.720

6"are not completely metrizable" is the wording you want. – Andrés E. Caicedo – 2017-07-15T03:28:48.397

3@KConrad and Andres: The wording is part of what made that false belief so believable! At the time I didn't think about the fact that there could be multiple metrics, much less that the completeness of those metrics wasn't a topological property. I only realized my mistake when I was introduced to the ideas contained in your two comments. (That, and picturing the irrationals as complete is HARD!) – Pace Nielsen – 2017-07-15T04:15:00.843

Pace, you probably realize by now that continued fractions make the picture a lot easier (whereby the space of irrationals between $0$ and $1$ can be identified with a product space $\mathbb{N}^\mathbb{N}$). – Todd Trimble – 2017-07-23T21:07:36.573


@DuchampGérardH.E. A topological subspace of a completely metrizable topological space is completely metrizable if and only if it is a $G\delta$, that is a countable intersection of open sets. One can use Baire's category theorem to show that $\mathbb{Q}$ is not a $G\delta$. All this can be found at: https://en.wikipedia.org/wiki/G%CE%B4_set

– Michael Greinecker – 2017-08-11T14:41:37.623

@MichaelGreinecker Then, The irrational numbers, in their usual topology as a subset of $\mathbb{R}$, is a completely metrizable space. This is a consequence of Mazurkiewicz's theorem, am I right ? (thank you for the reference +1). – Duchamp Gérard H. E. – 2017-08-11T16:58:32.023

1@DuchampGérardH.E. Exactly, it is the countable intersection $\bigcap_{q\in\mathbb{Q}}\mathbb{R}\setminus{q}$ of open sets. – Michael Greinecker – 2017-08-11T17:05:32.767


These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.


Posted 2010-05-04T21:02:58.510

Reputation: 1 992

What "people"? Non-mathematicians? – Todd Trimble – 2011-05-04T00:03:41.390

1@Todd: No i was talking of high school students. – crskhr – 2011-05-04T04:08:55.983

@S.C.:if squarring both sides will not give the solution then how can second problem be solved? – PK Styles – 2017-10-25T16:08:05.263


Hopefully this isn't a repeat answer. False belief: a matrix is positive definite if its determinant is positive.

Aryeh Kontorovich

Posted 2010-05-04T21:02:58.510

Reputation: 1 715

6Is this really a common(!) false belief? – Martin Brandenburg – 2011-10-03T07:23:59.563


The "conditional Vitali convergence theorem": Let $X_n$ be a uniformly integrable sequence of random variables with $X_n \to X$ almost surely, and $\mathcal{G}$ a sub-$\sigma$-field. Then $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ almost surely (FALSE).

I believed this one until I read Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute. It is particularly startling because the conditional versions of the monotone convergence theorem, the dominated convergence theorem, and Fatou's lemma are all true!

What is true is that $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ in $L^1$, so you do have a subsequence converging almost surely.

Nate Eldredge

Posted 2010-05-04T21:02:58.510

Reputation: 17 394


In ${\mathbb F}_p^\times$, the non-squares are the opposite of the squares. In other words, $a$ is square iff $-a$ is not a square.

This is a confusion with the facts that the kernel of $x\mapsto x^2$ is $\{1,-1\}$ and the subgroup of squares has index $2$.

Denis Serre

Posted 2010-05-04T21:02:58.510

Reputation: 26 530


Here is a false belief I had. Let $f:X \to Y$ be a map of topological spaces having the property that for every finite CW complex $K$, the induced map $f_{\ast}:[K,X] \to [K,Y]$, on unpointed homotopy classes of maps, is a bijection. Then $f$ is a weak homotopy equivalence (that is, it induces isomorphisms on all homotopy groups relative to all basepoints). A counterexample is given by the stabilization map $B \Sigma_{\infty}\xrightarrow{+1} B \Sigma_{\infty}$, which is not an isomorphism on $\pi_1$.

Ilan Barnea

Posted 2010-05-04T21:02:58.510

Reputation: 889

7Although the original intent of this question seems to have long since evaporated, I can't help asking: is this really a "common false belief"? – Yemon Choi – 2015-02-17T01:24:00.800

how about: if two CW complexes have all homotopy groups isomorphic, then they are homotopy equivalent? as i recall, you need those isomorphisms to be induced by a single continuous map. – roy smith – 2017-04-22T00:01:46.037

@roysmith Yes. You can even have two non weakly equivalent spaces having all Postnikov stages weakly equivalent – Ilan Barnea – 2017-05-08T10:47:36.133


A few mistakes I remember:

  • The quotient groups $\frac{G}{N}$ and $\frac{H}{K}$ are isomorphic if $G \thicksim H$ and $N\thicksim K$.
  • A closed interval of a complete lattice is a complete sublattice.
  • Two homeomorphic topologies on a set are the same.
  • The set of all compatible uniformities of a topological group forms a complete lattice.
  • The trace of the identity matrix is 1.


Posted 2010-05-04T21:02:58.510

Reputation: 229

2A closed interval of a complete lattice does form a lattice that is complete, right? So that the mistake is that sups and infs in the interval (particularly the sup and inf over the empty set) are not necessarily computed as they would be in the ambient complete lattice; is that what you have in mind? – Todd Trimble – 2015-09-06T01:47:02.330

Yes‌‌‌‌‌‌‌‌‌‌‌‌. – user47958 – 2015-09-06T01:55:38.963


Most people that study Riemannian geometry for their first time make the following assumption at some point: "Let $(e_1,\dots,e_n)$ be a local orthonormal frame of $TM$ such that all Lie brackets $[e_i,e_j]$ vanish..."

This one is not so common (maybe special to me), but here we go: "$\mathbb{RP}^\infty$ and $\mathbb{CP}^\infty$ are Eilenberg-Mac Lane spaces, so $\mathbb{HP}^\infty$ is one, too."

Sebastian Goette

Posted 2010-05-04T21:02:58.510

Reputation: 4 783


(1) All Lebesgue-null sets are countable, or are strongly measure zero. (2) The following,verbatim, was a Q in American Mathematical Monthly : " A student asserted that any uncountable real set has a closed uncountable subset. Is this true ?" .


Posted 2010-05-04T21:02:58.510

Reputation: 180


This example is similar to this earlier answer.

If $k$ is a field, then $k[x] \otimes_k k[y] \cong k[x,y]$. Therefore also $k[[x]] \otimes_k k[[y]] \cong k[[x,y]]$, right?

R. van Dobben de Bruyn

Posted 2010-05-04T21:02:58.510

Reputation: 7 200


"This algebraic variety is a $C^\infty$-smooth manifold, therefore it must be non-singular". This sounds obvious (and in fact it is true over $\mathbb{C}$) however it is false in general (for instance over $\mathbb{R}$). See the discussion here for many details.

A. S.

Posted 2010-05-04T21:02:58.510

Reputation: 21


"It cannot be shown without some form of AC that the union (or disjoint union) of countably many countable sets is countable. I have a countably infinite set X of countably infinite sets. Therefore, the union of X cannot be shown to be countable without Choice."

The fallacy is that in many cases of interest, it is possible to exhibit an explicit counting of every element of X. In such a case a counting of X by antidiagonals is easily constructed. The usual counting of the rationals is an example of this.

I think this may even be an example of a more general phenomenon of "people think AC is necessary for a certain construction, but in fact it turns out not to be necessary for the example they have in mind". For example, AC is necessary to find a maximal ideal in an arbitrary ring ... but it isn't if you're prepared to assume the ring is Noetherian.


Posted 2010-05-04T21:02:58.510

Reputation: 1

3If "Noetherian" is defined by the ascending chain condition or by requiring all ideals to be finitely generated, then in order to deduce the existence of maximal ideals, you still need a weak form of the axiom of choice. The usual argument uses the axiom of dependent choice. (Of course, if you define "Noetherian" to mean that every set of ideals has a maximal element, then deducing the existence of maximal ideals is a choiceless triviality.) A good reference is "Six impossible rings" by Wilfrid Hodges (J. Algebra 31 (1974) 218-244). – Andreas Blass – 2010-10-22T15:29:47.227

Thanks Andreas! I had a feeling there was a technicality somewhere there, but couldn't remember what it was. As a philosophical point I personally think that of course in the absence of AC you want to define Noetherian so that my original statement is true, but admittedly that's a harder sell than my countable-sets example. – Karol – 2010-11-16T21:06:15.710


If $a$ is a real zero of a cubic polynomial with rational coefficients then $a$ can be written as a combination of cube roots of rational numbers.

More generally if $a$ is a real zero of an irreducible polynomial with rational coefficients that is solvable by radicals then students expect the following:

  1. Any expression inside a radical evaluates to a real number
  2. Any sub-expression of the expression for $a$ evaluates to an algebraic number of order less than or equal to the order of $a$

Of course the problem is that from Cardan's solution to the cubic we can have negative rational numbers inside a square root. Let $c$ = $4*(-1 + \sqrt{-3})$.

$a$ = $\frac{\sqrt[3]{c}}{4} + \frac{1}{\sqrt[3]{c}}$

$f(x) = 4x^3 - 3x + \frac{1}{2}$.

So while $a$ is an algebraic number of degree three, it can not be written as combination of cube roots of rational numbers. Indeed, it is counter-intuitive that $\sqrt[3]{c}$ has degree 6 over the rational numbers yet we can use this number and simple arithmetic to produce an algebraic number of degree 3.

Also $a$ = $\sin(50^{\circ})$. For many values of $\theta$, $\sin \theta$ is a radical number. See also radical values for sine and cosine

Greg Gibson

Posted 2010-05-04T21:02:58.510

Reputation: 143


This has a name: http://en.wikipedia.org/wiki/Casus_irreducibilis

– Qiaochu Yuan – 2011-04-06T21:21:18.747


Coordinates on a manifold do not have an immediate metric meaning. Until becoming familiar with differential geometry one tends to think they do. (Einstein wrote that he took seven years to free himself from this idea.)

For example, linear control theory is for the most part metric with variables in $R^n$. When moving away from linear control theory, variables are represented as coordinates on a manifold. Nevertheless, much of the literature tends to either abandon metric notions altogether, or to keep using an Euclidean metric though it is no longer very useful.


Posted 2010-05-04T21:02:58.510

Reputation: 508


Given a bundle $ E \to X$, let $\mathcal{E}$ denote its sheaf of sections.

False belief: Given a map $f: Y \to X$, the inverse image sheaf $f^{-1}\mathcal{E}$ is the sheaf of sections of the pullback bundle $f^* E \to X$.

This is true if $E \to X$ is a local homeomorphism (e.g. a covering space), or if $f: U \hookrightarrow X$ is the inclusion of an open subset, but not for general maps and bundles.

For instance, taking $x^{-1}\mathcal{E}= \mathcal{E}_x$ for $x: 1 \to X$ the inclusion of a point and $\mathcal{E}$ the sheaf of smooth functions on a manifold will demonstrate that it is false.

For vector bundles (or sheaves of modules over the structure sheaf of a ringed space in general), the correct statement is obtained by using the pullback functor $$f^*\mathcal{V} = \mathcal{O}_Y \otimes_{f^{-1}\mathcal{O}_X} f^{-1} \mathcal{V}$$ which is the inverse image followed by extension of scalars.

One issue which leads to this false belief is that texts on sheaves often use $f^*$ in place of $f^{-1}$ for the inverse image functor, rather than reserving the former for sheaves of modules over ringed spaces.


Posted 2010-05-04T21:02:58.510

Reputation: 2 322


False belief: It is obvious how to prove that $\sin'=\cos$.

Not so much... if $\cos$ and $\sin$ are defined geometrically. You need to prove geometrically that $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ and a (non-circular) proof of that is not obvious (see here).

Personally I'm aware of that just today! (thanks to a remedial course given to my niece).

Sebastien Palcoux

Posted 2010-05-04T21:02:58.510

Reputation: 7 117

1The string of comments below the Math.SE link shows that really pursuing this gets one down a rabbit hole of rigorous discussions of arc length and such. On the other hand, one can prove that if $S$ (think "sine") and $C$ (think "cosine") are continuous functions which satisfy the standard addition formulas and the Pythagorean theorem $C^2 + S^2 \equiv 1$, then $S'(0)$ exists and $S'(x) = S'(0)C(x)$. There is a whole family of sine-like functions $S_a: x \mapsto S(ax)$; adjusting the parameter $a$ so that $S_a^\prime(0) = 1$, you can define the standard sine to be this $S_a$: a neat finesse. – Todd Trimble – 2017-10-08T03:14:08.157

1@ToddTrimble: Should we interpret your comment as for or against this false belief? Or else, is it just a neutral complement? For a full proof, is your way really easier? – Sebastien Palcoux – 2017-10-10T11:56:21.160


It's mainly in agreement with you, with a shade of neutral. I don't claim that my suggestion really makes it easier, but only that one can prove things rigorously without getting into considerations of arc length, with (theoretically anyway) no prerequisites past introductory differential calculus. Mainly it's based on convexity arguments; I have a write-up here: https://ncatlab.org/toddtrimble/published/Continuity+of+the+exponential+function; see theorem 3.1 and the crucial lemma 3.4. The "finesse" is akin to how we adjust parameter $a$ in $f: x \mapsto a^x$ to $a = e$ to get $f'(0) = 1$.

– Todd Trimble – 2017-10-10T13:22:09.163


Heisenberg-Weyl and enveloping algebras

I have heard (and have read) the following belief even among distinguished mathematical physicists.

Let $HW_\mathbb{C}$ be the (associative with unit) algebra generated by two elements $\{a,a^\dagger\}$ subjected to the relation $[a,a^\dagger]=1$. It is easy to check that $$ \mathfrak{g}=span_\mathbb{C}\{a,a^\dagger,1_{HW_\mathbb{C}}\} $$ is a Lie subalgebra.

False belief The algebra $HW_\mathbb{C}$ is the universal enveloping algebra of $\mathfrak{g}$ i.e. $$ HW_\mathbb{C}=U(\mathfrak{g})\ . $$

One way to see that this is false at once is to observe that any enveloping algebra $U(\mathfrak{g})$ possesses (at least) a character $\varepsilon$, but there is none on $HW_\mathbb{C}$ (one would have indeed $1=\varepsilon([a,a^\dagger])=0$).

Duchamp Gérard H. E.

Posted 2010-05-04T21:02:58.510

Reputation: 2 035

1Yeah. There is a surjection $U\left(\mathfrak{g}\right) \to HW{\mathbb{C}}$, though; its kernel is the ideal generated by $1{HW{\mathbb{C}}} - 1$. Thus, $HW{\mathbb{C}}$ can be viewed as a "reduced" $U\left(\mathfrak{g}\right)$ (similarly to how the free product $A * B$ of two algebras $A$ and $B$ can be viewed as a "reduced" tensor algebra $T\left(A \otimes B\right)$). This is one of the things Pavel Etingof taught me; before that, I thought Weyl/Clifford algebras and universal enveloping algebras were similar but un-relatable constructions. – darij grinberg – 2017-10-03T06:34:30.593

You're right ! the enveloping algebra of $\mathfrak{g}=span\mathbb{C}{a,a^\dagger,1{HW\mathbb{C}}}$ is the algebra (associative with unit) generated by ${a,a^\dagger,e}$ ($e$ is a clone of $1{HW_\mathbb{C}}$) subjected to $$ [a,a^\dagger]=e\ ;\ [a,e]=[a^\dagger,e]=0 $$ combinatorially, in the reduced expressions, the power of $e$ counts the number of times Wick commutations have been applied. I'd include this in my post if it were not out of place. – Duchamp Gérard H. E. – 2017-10-03T12:37:31.433


This may not count as a false belief, but it is an amusing misconception I had. In college I had a numerical analysis professor who had both a strong accent and messy handwriting, so it was hard to know exactly what he was talking about sometimes.

I was not yet familiar with the Greek letter $\xi$, and that was the variable he always used to represent the error of a computation, but with his handwriting it just looked like a purposeful scribble. So he would say,"Here we have the calculated value and then of course with some error" (scribble).

I thought he was just being dismissive about the error and trying to represent it in a pejorative way.

To be fair, the letter $\xi$ is not one of the easier ones to draw by hand.

Somatic Custard

Posted 2010-05-04T21:02:58.510

Reputation: 197

3One of my high school teachers was known for exclaiming "I'm a genius!" in reference to certain multivariate polynomials. – Dan Brumleve – 2018-01-13T17:46:43.050


I used to think that the subset of even norm vectors in an integral lattice is a sub-lattice. This is true for the "classically integral" lattice defined by $<u,v> \in \mathbb{Z}$ for $u,v$ in the lattice because the even vectors is the kernel of the group homomorphism $v \rightarrow <v,v>$ mod 2. However this fails for the more general notion of "integer norm" lattice where we only require the quadratic form is integer valued (ie. the coefficients are all integral or that the off diagonal entries in the Gram matrix may be half integral eg $x^2+xy+2y^2$). For the hexagonal lattice $x^2+xy+y^2$ which is not classically integral, the even vectors is a sub-lattice but for a different reason that it is the lattice scaled by 2.

If $t = 3$ mod 4, the lattice $L_t$ with quadratic form $x^2+ty^2$ is classically integral. Its even sublattice $L_{t0}$ has quadratic form $4(x^2+xy+(t+1)y^2/4)$ which clearly equals 2$W_t$ where $W_t$ is the lattice with quadratic form $(x^2+xy+(t+1)y^2/4)$. If $t=7$ mod 8, the coefficients of the form is [odd,odd,even], the even vectors is not a subgroup since for example $[0,1]$ and $[1,1]$ has even norm but $[0,1]+[1,1]=[1,2]$ has odd norm. If $t$ is 3 mod 8, the form $(x^2+xy+(t+1)y^2/4)$ can only be even only if both $x,y$ are even since all coefficients are odd. So the even vectors in $W_t$ turn out to be $2W_t$. It is a sub-lattice and it is the subset of even vectors but it is index 4 in $W_t$. It is the 2-scaled sub-lattice.

If $v \in L_t$, $2v \in 2L_t \subset L_{t0}=2W_t$, so $L_t \subset W_t$. So the picture is $2W_t=L_{t0} \subset L_t \subset W_t$ with each containment is index 2.


Posted 2010-05-04T21:02:58.510

Reputation: 76


A somewhat common belief among students starting out in cryptography:

Breaking RSA requires factoring the modulus.

Although it is not quite known to be a "false" belief, there is no known reduction showing that breaking RSA implies finding the prime factors of the modulus. This in contrast to e.g. Rabin's cryptosystem, and various cryptographic schemes built on other hard problems, whose security provably relies on the underlying hard problems.


Posted 2010-05-04T21:02:58.510

Reputation: 273


I would like to turn the attention of mathematical community to a false beliefs related to the direct limit topologies.

Many years ago in the theory of topological groups there was a false belief that for every space $X$ the free topological group carries the topology of direct limit of the sequence $F_n(X)$ of words of length $\le n$. This illusion was broken up by Fay, Ordman and Thomas who showed that even for the space of rational numbers the free topological group $F(\mathbb Q)$ is not a $k$-space.

The problems with direct limit topologies is that for the direct limit $X=lim X_n$ of an increasing sequence $(X_n)$ of topological spaces the topology on $X\times X$ does not coincide with the direct limit topology of the sequence $ (X_n\times X_n)$.

Now specialists in General Topology and Topological Algebra are conscious of pathological behaviour of direct limit topologies and are careful with this delicate topic.

On the other hand, I was quite surprised lerning that in Algebraic Geometry this misbelief still is alive. For example, in this paper posted to arxiv (maybe it is already published) in the very introduction (on page 3) it is written that for any topological space $X$ the Ran space (of all non-empty finite subsets of $X$, endowed with the topology of direct limit of the sequence $R_n(X)$ of sets of cardinality $\le n$ in $X$) is a topological semilattice. But this is not true in general, see Proposition 4 here.

So, some false beliefs that have died in some areas of mathematics can be still alive in others. By the way this sutiation also explain why mathematicians should not neglect general topology.

Taras Banakh

Posted 2010-05-04T21:02:58.510

Reputation: 8 618


False Belief: "The suspension spectrum map from spaces to (edit: symmetric) spectra preserves smash-products"

The facts that one denotes the smash product of spectra and the smash product of a space with a spectrum (levelwise) with the same $\wedge$ and tends to leave away the $\Sigma^\infty$ when one embeds a space into spectra are also not helpful in getting used to the harsh reality that the above is wrong.

Peter Arndt

Posted 2010-05-04T21:02:58.510

Reputation: 8 494

Yay! 100th answer! – Peter Arndt – 2010-10-04T21:33:27.240

4I don't see that this qualifies as a false belief. In order for the question of whether it is true or false to even be meaningful, you have to first commit yourself to one of the many different notions of spectrum, not to mention smash product of spectra. – Tom Goodwillie – 2010-10-05T00:35:50.970

1True. I meant symmetric spectra with the smash product coming from their description as modules over the symmetric sequence of spheres. – Peter Arndt – 2010-10-05T10:52:19.667


If a matrix $A$ is self-adjoint/skew-self-adjoint with respect to a symmetric bilinear form, then it is diagonalizable.

True for matrices over $\mathbb{R}$, with respect to a positive definite inner product.

False over other fields. For example, over $\mathbb{C}$, $\left( \begin{smallmatrix} 1 & i \\ i & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & 1 & i \\ -1 & 0 & 0 \\ -i & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to the standard inner product.

False for other nondegenerate symmetric bilinear forms: $\left( \begin{smallmatrix} 1 & 1 \\ -1 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to $\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{smallmatrix} \right)$.

You can exponentiate the skew-self-adjoint matrices to get examples of matrices preserving a nondegenerate symmetric bilinear form, with Jordan blocks of the form $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

David E Speyer

Posted 2010-05-04T21:02:58.510

Reputation: 98 510

7You seem to have a different definition of "the standard inner product on $\mathbb{C}^n$" than I do. I think that phrase normally refers to the familiar positive definite sesquilinear form, with respect to which self-adjoint matrices are indeed diagonalizable. – Mark Meckes – 2011-01-28T16:46:14.937

But that's not a bilinear form. And it has no generalization to other fields (what is it on $\overline{\mathbb{F}_p}$?). How can it be standard? :) I certainly agree that people should know that matrices which are self-adjoint with respect to the standard sesquilinear form are diagonalizable. – David E Speyer – 2011-01-28T17:51:03.517

6Of course it's not bilinear -- an "inner product" on a complex vector space is defined to be sesquilinear, not bilinear -- I've spent a lot of time trying to get my linear algebra students to remember that. The failure of such a form to generalize to other fields is indeed sad, but I think the richness of Hilbert space theory helps to make up for that disappointment. :) – Mark Meckes – 2011-01-28T21:24:50.017


Something I was sure about until earlier today:

Suppose $\kappa$ is an $\aleph$ number, then $AC_\kappa$ is equivalent to $W_\kappa$, namely the universe holds that the product of $\kappa$ many sets is non-empty if and only if every cardinality is either of size less than $\kappa$ or has a subset of cardinality $\kappa$.

In fact this is only true if you assume full $AC$, and $(\forall \kappa) AC_\kappa$ doesn't even imply $W_{\aleph_1}$, I was truly shocked.

Furthermore, $W_\kappa$ doesn't even imply $AC_\kappa$ in most cases.

The strongest psychological implication is that most people actually think of the well-ordering principle as a the "correct form" of choice, when it is actually Dependent Choice (limited to $\kappa$, or unbounded) which is the "proper" form, that is $DC_\kappa$ implies both $AC_\kappa$ and $W_\kappa$.

Asaf Karagila

Posted 2010-05-04T21:02:58.510

Reputation: 18 539

8How common is this misconception? – Thierry Zell – 2011-04-17T03:08:08.420

3@Thierry: For the past couple of weeks I spent a lot time considering models without choice, not only I held that misconception but not once anyone corrected me about it - grad students and professors alike. – Asaf Karagila – 2011-04-17T06:09:35.543


Assume that $a,b\in \mathbb{R}\setminus \{0\}$ which satisfy $a^{3}= 2b^{3}$.

Then $a-2b$ is a non zero nilpotent element of group ring $\mathbb{Z}_{3} \mathbb{R}$, that is $(a-2b)^{3}=0$.

This would be a counterexample to the zero divisor Kaplansky conjecture

The false lies in an obvious abuse in the definition of the group ring multiplication.

Ali Taghavi

Posted 2010-05-04T21:02:58.510

Reputation: 234

3This does not seem like a common false belief. – Yemon Choi – 2016-05-15T11:47:36.347


I once misunderstood the definition of monads, and thought that for a monad $(T,\eta,\mu)$, we have $T\eta_X = \eta_{TX}$ (or fmap return == return in Haskell). Of course this is not the case (in case of $T=$[], fmap return [1,2] is [[1],[2]], whereas return [1,2] is [[1,2]]).

H Koba

Posted 2010-05-04T21:02:58.510

Reputation: 89


Let $M \subset B(H)$ be a von Neumann algebra, $p \in B(H)$ a projection and $q=I-p$.

False belief: If $pM=Mp$ then $M=pMp \oplus qMq$.
(I think it is a quite common careless mistake)

Counter-example: diagonal embedding of $\mathbb{C}$ into $M_2(\mathbb{C})$.

Sebastien Palcoux

Posted 2010-05-04T21:02:58.510

Reputation: 7 117


I don't think I've seen it in here:

Every vector space has a non-trivial dual space ($L^p$ for $0 < p < 1$ was a counter-example only mentioned during one of the classes in measure theory)

And of course there's the common false belief of people outside of mathematics that "mathematicians work with numbers and formulae all day long" :)

Asaf Karagila

Posted 2010-05-04T21:02:58.510

Reputation: 18 539

11Well, it is true that every vector space has a dual space, even $L^{1/2}$... and it is even true that every topological vector space has a continuous dual space... What you mean is that it is not true that every topological vector space has a non-trivial continuous dual space (or, that the continuous dual of a topological vector space does not necessarily separate points) – Mariano Suárez-Álvarez – 2010-07-07T18:54:26.430

You are indeed correct. I'll do better not to dismiss the trivial case the next time. – Asaf Karagila – 2010-07-07T19:31:04.080


Fans: (related to the one of polytopes written above) all convex cones are rational, i.e. one would expect that a line would eventually hit a point in the lattice. It is obviously not true, just take the one-dimensional cone generated by $(1,\sqrt{2})$. A similar one was thinking that if I rotate the cone a bit, I can always make it rational.

Jesus Martinez Garcia

Posted 2010-05-04T21:02:58.510

Reputation: 993

3reminds me of the curious fact that some circles in the plane, too, have no points in $\mathbb Q^2$. (proven simply by cardinality!) – AndrewLMarshall – 2010-10-04T19:21:50.673


The assumption that a cubic surface expressed as a foliation of Weierstrass curves cannot be rational, because a general Weierstrass curve is not rational.

I've seen this false assumption more than once on sci.math over the years. But there are simple counterexamples, such as:

$ (x + y) (x^2 + y^2) = z^2 $

On defining $ u = x/y $ and $ v = z/y $ one obtains $ y (u + 1) (u^2 + 1) = v^2 $, and hence x, y, z as rational functions of u, v.

I'd love to have a reference to a procedure for calculating the geometric genus and algebraic genus of surfaces like this, because they are rational if and only if both these quantities are zero, and for other cubic surfaces that interest me it would save a lot of fruitless hacking around trying to find a rational solution that probably doesn't exist! Are there any symbolic algebra packages that can do this?

I mean for example is $ x y (x y + 1) (x + y) = z^2 $ rational? I'm almost sure it isn't; but how can one be sure?

John R Ramsden

Posted 2010-05-04T21:02:58.510

Reputation: 836

Algebraic genus in software: Seems to be available in Singular and Maple and may be available in Mathematica by now (from https://mathematica.stackexchange.com/a/5453/16237 ). (Aside: I don't know anything about the software mentioned.)

– Eric Towers – 2017-08-13T02:06:03.030


This might not be common, but I once believed the following.

Let $ A, B $ be integers, and define a sequence by the linear recurrence $ s_n = A s_{n-1} + B s_{n-2} $ with the base case $ s_0 = 0 $, $ s_1 = 1 $. Two important special cases are the Fibonacci sequence ($ A = B = 1 $) and the sequence $ s_n = 2^n - 1 $ (where $ A = 3 $, $ B = -2 $). Then, for any integers $ n $ and $ k $, $ \gcd(s_n, s_k) = s_{\gcd(n,k)} $.

This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $ A = B = k = 2 $, $ n = 3 $.

Update: corrected the powers of two minus one example from $B = 2$ to $B = -2$. Thanks to Harry Altman.

Zsbán Ambrus

Posted 2010-05-04T21:02:58.510

Reputation: 1 191

Quick correction, that should be A=3, B=-2 for 2^n-1. – Harry Altman – 2011-04-07T21:23:47.587


If every collection of disjoint open sets in a topological space is at most countable, then the space is separable


Posted 2010-05-04T21:02:58.510

Reputation: 301


This might not be common, but it gave me a headache once. I'll delete if it gets heavily downvoted.

I once had to think really hard about a contradiction in the great scheme of things that followed from my unwitting assumption that if $f$ was a function from a semigroup to a semigroup, then if its kernel was a congruence, $f$ had to be a homomorphism. I encountered a function whose kernel clearly was a congruence but which clearly wasn't a homomorphism, and it took about an hour's walk in a park for my vague notions and incoherent thought to produce the necessary realization.

Michał Masny

Posted 2010-05-04T21:02:58.510

Reputation: 437


A common false assumption is that that two non-orthogonal pure states of a quantum mechanical system may never be unambiguously distinguished by a measurement. (See http://arxiv.org/pdf/quant-ph/9807022.pdf)

Another false belief is that a quantum computer is similar to an analogue computer, in that large computations will necessarily fail because of accumulated error. (See, for example, http://arxiv.org/abs/quant-ph/9712048)

For that matter, another common false believe is that Bell Inequalities aren't violated, although it is mostly held by people who have never heard of Bell Inequalities.

J Tyson

Posted 2010-05-04T21:02:58.510

Reputation: 71

3I'm not sure how you can believe that something you have never heard of isn't violated. – Geoff Robinson – 2016-04-10T17:55:42.093


Let $R$ be a ring with identity $e$, $A, B\in R$, $A\neq 0$, $B$ is invertible element. If $A\cdot B = A$ then $B = e$.

Mikhail Goltvanitsa

Posted 2010-05-04T21:02:58.510

Reputation: 359


Another common mistake. If $W = _P(e_1,\ldots, e_{n})$ is a vector space and $V$ is a subspace of $W$ of dimension $k$, then $V = _P(e_{i_1},\ldots, e_{i_k})$.

Mikhail Goltvanitsa

Posted 2010-05-04T21:02:58.510

Reputation: 359

What does that little subscript $p$ on the equals sign mean? – Gerry Myerson – 2016-02-11T21:36:46.247

$V$ is a vector space over field $P$. – Mikhail Goltvanitsa – 2016-02-12T06:52:15.480

So, what does "$V$ is a vector space over field $P$ $(e_1,\dots,e_n)$" mean? – Gerry Myerson – 2016-02-12T08:37:51.120

1$W$ is a vector space over field $P$, $(e_1,\ldots, e_n)$ is a basis of $W$. $V$ is a subspace of $W$. – Mikhail Goltvanitsa – 2016-02-12T08:40:25.007


Here's one that I think will surprise many.

False belief. Let $E$ be an elliptic curve over an algebraically closed field $k$ of characteristic $p > 0$. Then $\operatorname{End}^\circ(E)$ is strictly larger than $\mathbb Q$.

While this is true for all elliptic curves defined over finite fields, most elliptic curves whose field of definition is transcendental over $\mathbb F_p$ have $\operatorname{End}^\circ(E) \cong \mathbb Q$. The extra automorphism on elliptic curves over a finite field comes from the geometric Frobenius. For varieties over larger fields, this is not a thing.

R. van Dobben de Bruyn

Posted 2010-05-04T21:02:58.510

Reputation: 7 200


People seem to believe that conventional computation (for example, running a chaotic irreversible cellular automaton) can be as efficient as one wants simply with good engineering, but this is not the case. Landauer's principle states that erasing a bit of information always takes $\ln(2)\cdot k\cdot T$ energy where $k$ is Boltzmann's constant ($k=1.38065\cdot 10^{-23}$ Joules/Kelvin) and $T$ is the temperature. Landauer's principle is a consequence of the second law of thermodynamics since if Landauer's principle were violated, then entropy would decrease. Landauer's principle means that conventional irreversible computation always must take $\ln(2)\cdot k\cdot T$ energy per bit erased (and one can erase data just by running it through AND and OR gates, so every irreversible gate must take a minimum amount of energy by Landauer's principle). However, Landauer's principle does not apply to reversible computation since reversible computers are not allowed to erase data.


Posted 2010-05-04T21:02:58.510

Reputation: 1

2Okay, I see what you are aiming at with this last edit. The idea that ordinary computation can be made arbitrarily efficient is a reasonably common false belief about our physical world (and may even have a somewhat solid mathematical interpretation). I withdraw my initial objection. – S. Carnahan – 2017-08-11T14:04:39.033


False belief: << Let $M$ be the von Neumann algebra generated by a $\rm{C}^{\star}$-algebra $\mathcal{A}$. >>

The false belief is to think that the above sentence makes sense. In fact, a von Neumann algebras and a $\rm{C}^{\star}$-algebra don't have the same status. A von Neumann algebra is an operator algebra by definition, i.e. it is defined inside $B(H)$ for some separable Hilbert space $H$. Now, some subalgebras of $B(H)$ are (separable) $\rm{C}^{\star}$-algebras, but a $\rm{C}^{\star}$-algebra can also be defined abstractly. It can next be representated and a given representation $H$ (defined for example by GNS construction for a given state), if it is faithful, induces an embedding in $B(H)$.
So to make sense, the sentence above should be modified as:

<< Let $M$ be the von Neumann algebra generated by $(\mathcal{A},\rho)$, a couple of $\rm{C}^{\star}$-algebra and state. >>
<< Let $M$ be the von Neumann algebra generated by a $\rm{C}^{\star}$-algebra $\mathcal{A}$ represented on $H$. >>

Then, $M = \pi_H(\mathcal{A})''$. We can use $M$ to characterize the representation $H$, for example, we can talk about a representation of type ${\rm I}$, ${\rm II}$ or ${\rm III}$ if $M$ is a von Neumann algebra of type ${\rm I}$, ${\rm II}$ or ${\rm III}$. There is a $\rm{C}^{\star}$-algebra with representations of every type, for example the Cuntz algebra.

Finally, there exists a universal representation for every $\rm{C}^{\star}$-algebra (i.e. the direct sum of the corresponding GNS representations of all states; it is faithful). The associated von Neumann algebra is called the enveloping von Neumann algebra (it can also be defined as the double dual); it contains all the operator-algebraic information about the given $\rm{C}^{\star}$-algebra.

Sebastien Palcoux

Posted 2010-05-04T21:02:58.510

Reputation: 7 117

So there is no abstract version of the notion of a von Neumann algebra? Like, say, isomorphism classes of "usual" von Neumann algebras, or something like that? – მამუკა ჯიბლაძე – 2018-01-12T21:53:37.660


@მამუკაჯიბლაძე: A von Neumann algebra can be defined abstractly as a (non-necessarily separable) $\rm{C}^{\star}$-algebra that have a predual; but it is not the usual definition, some authors call this abstract version a $\rm{W}^{\star}$-algebra, see the last paragraph of https://en.wikipedia.org/wiki/Von_Neumann_algebra#Definitions

– Sebastien Palcoux – 2018-01-12T22:23:08.657


False belief: relativization is well-defined and the corresponding notation $C^A$ is unambiguous. Which is not quite true because $P=NP$ would not imply $P^A=NP^A$.


Posted 2010-05-04T21:02:58.510

Reputation: 1 055

1Maybe some more explanation would be useful. If decision problems and oracles are subsets of $\mathbb{N}$, and complexity classes are subsets of $P(\mathbb{N})$, then there is in general no such operation as relativization. I'm not sure how common of a false belief this is, but once I settled on my preference for the set point of view and saw what was going on here I lost some interest in the idea of relativization. – Dan Brumleve – 2018-01-13T05:16:39.750


For $p$ prime and the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \cdots$ given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is not the group of $p$-adic integers $\mathbb{Z}_p$, but its Pontryagin dual, the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$.

Sebastien Palcoux

Posted 2010-05-04T21:02:58.510

Reputation: 7 117

2Is that actually a common false belief? After all, $\mathbb{Z}_p$ is uncountable, as everyone realizes! – Todd Trimble – 2015-03-05T14:25:20.433

"$\mathbb{Z}_p$ is countable" is also a false belief for people who didn't really read the definition of $\mathbb{Z}_p$, but I don't know how much it is common. – Sebastien Palcoux – 2015-03-05T14:34:37.010

6It's hard for me to believe it's at all common. I wasn't the downvoter, but I think it would be better if answers were rooted either in instances that can be found in the literature, or widely encountered in one's experience as an instructor. – Todd Trimble – 2015-03-05T14:52:11.870


I don't know how common this is, but it occurs as a corollary of a theorem in the fine, and widely used, text by Shafarevich on algebraic geometry: namely, if $f \colon X \longrightarrow Y$ is a surjective algebraic map of varieties, then 1) for all $y \in Y$, the fiber over $y$ has dimension $≥ \dim(X)-\dim(Y)$; 2) on some non empty open set in $Y$ the dimension of the fibers equals $\dim(X)-\dim(Y)$; 3) for all $r$, the set of $y \in Y$ such that the fiber over $y$ has dimension $≥ r$, is closed in $Y$.

The first two are true, but the third is false. Upper semicontinuity of fiber dimension is true on the source, not the target. For the conclusion as stated to hold, one can add properness to the hypothesis on the map. I think this is not at all widely believed by experts, but for some reason it persists in the text, hence may be believed by students.

Since I have myself written notes in which blatantly false statements occur, I do not think for a moment that Shafarevich himself believed this false statement. But such things do slip by, and may mislead beginners. In fact I believed it for some time until enlightened by a friend.

In keeping with the OP's desire to know the psychological reason for the error, it seems for some reason common in my experience for people to assume unconsciously that maps are proper.

roy smith

Posted 2010-05-04T21:02:58.510

Reputation: 8 641


I checked many answers but did not find this one. Probably I missed it.

Let $G$ be reductive group over a p-adic field, $B$ be its Borel subgroup and $w_\ell$ be its long Weyl element.

False belief: For a smooth compact supported function $f$ on $G$, then $f|_{B w_\ell B}$ is also smooth compact supported function on $B w_\ell B$.

My advisor told me that this kind mistake is quite common for beginners.

Here is another subtle false belief. Still let $F$ be a p-adic field and let $\mathfrak{p}$ be its prime. Consider the metaplectic double cover $$\widetilde {\textrm{Sp}}_{2n}(F)\rightarrow \textrm{Sp}_{2n}(F).$$ Here as a set $\widetilde {\textrm{Sp}}_{2n}(F) =\textrm{Sp}_{2n}(F)\times \{\pm 1\}$, and the multiplication in $\widetilde{\textrm{Sp}}_{2n}(F)$ is define by Rao cocycle $c=c_{Rao}$: $(g_1,\epsilon_1)(g_2,\epsilon_2)=(g_1g_2,\epsilon_1\epsilon_2 c(g_1,g_2))$. Let $K_m^n=(1+M_{n\times n}(\mathfrak{p}^n))\cap \textrm{Sp}_{2n}(F)$ be the standard congruence subgroup.

False belief: The map from $K_m^n\rightarrow \widetilde {\textrm{Sp}}_{2n}(F),$ $k\mapsto (k,1)$ defines a group homomorphism if $m>>0$.

This is false even when $n=2$ and I saw this statement in published papers.

When $n=1$, this is correct. In general, the double cover $\widetilde {\textrm{Sp}}_{2n}(F)\rightarrow \textrm{Sp}_{2n}(F)$ splits over $K_m^n$ for $m$ large, but the naive map $k\mapsto (k,1)$ does not give the splitting for $n\ge 2$.

Qing Zhang

Posted 2010-05-04T21:02:58.510

Reputation: 295


"The topology of a unit ball in $\mathbb{R}^n$ is intrinsically wider than that of a lower-dimensional ball."

In fact, using a space-filling curve $[0,1]\rightarrow [0,1]^n$ one sees that the $n$-cube is a topological quotient of the interval.

Arnaud Mortier

Posted 2010-05-04T21:02:58.510

Reputation: 657

3Not sure I understand you, what's "wider"? – Michael – 2018-01-12T21:38:05.827

I don't refer to any mathematical notion there, just the common perception that there is "more room". – Arnaud Mortier – 2018-01-12T22:11:33.537


Way late to the party...

"$ \mathrm{polymod}\ p$ and $\mathrm{mod}\ p$ are the same thing."

And it's cousin: "$\forall{x}, f(x) \cong g(x) \pmod{q} \implies f(x) = g(x)$"


Posted 2010-05-04T21:02:58.510

Reputation: 1

11What does polymod mean? – darij grinberg – 2010-10-20T11:47:50.180

Either the cousin needs a bit more detail if it is to be false, it is quite naive! – Mariano Suárez-Álvarez – 2010-10-20T18:25:15.763


Probably I understand what this means: if $f(x)=0\pmod 2$ for all $x$, then $f=0$ over $\mathbb F_2$.

This is similar to my second example: http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/42919#42919

– zhoraster – 2010-10-20T18:33:06.220

Consequently, there are only $4$ polynomials over $\mathbb F_2$

Isn't this convenient? :-) – zhoraster – 2010-10-20T18:40:13.477

1$\mathrm{polymod}$ is "polynomial mod". Two polynomials are congruent $\mathrm{polymod} p$ iff the coefficients each power of the variable are congruent $\pmod{p}$. The equivalence classes are sets of polynomials where each coefficient ranges over an equivalence class $\pmod{p}$. For the cousin, there are many local/globals but they all seem to require additional conditions (q.v. Hensel lifting). I think the set from which $x$ was chosen was left unspecified because this "imprecise mental abbreviation" pops up at various levels of sophistication each with a different such set. – Anonymous – 2010-10-23T15:22:24.577


From Keith Devlin

"Multiplication is not the same as repeated addition", as put forward in Devlin's MAA column.

I'm not really sure how I feel about this one; I might be one of the unfortunate souls who are still prey to that delusion.


In case you missed it, the column ended up spilling a lot of electronic ink (as evidenced in this follow-up column), so I don't believe it would be wise to start yet a new one on MO. Thanks in advance!

Thierry Zell

Posted 2010-05-04T21:02:58.510

Reputation: 3 516

I followed your link, and I cannot even tell what is wrong about attaching helium balloons to both sides of a balance to model substraction on both sides of an equation. – user11235 – 2011-04-10T20:32:53.747

19The more I think about this "error", the less I am convinced.

It's like saying that you cannot say that $\binom n k$ is the number of $k$-element sets in an $n$-element set because then you will be unable to generalize to complex values of $n$.

Or you cannot define the chromatic polynomial as the function counting the colourings and then plug in $-1$ to get the acyclic orientations of the graph.

Also, I think it is perfectly understandable what it means to add something halfways. – user11235 – 2011-04-10T20:50:40.957


It's not a "false belief". It's a false heuristic. And it's actually here: http://mathoverflow.net/questions/2358/most-harmful-heuristic

– darij grinberg – 2011-04-10T21:17:48.327

@darij: one person's harmful heuristic is another person's false belief. I am sure that there are some people out there (admittedly not the most mathematically sophisticated) who sincerely believe that there is nothing more to multiplication than repeated addition. So I think this example also belongs here. – Thierry Zell – 2011-04-10T22:31:14.810

How do you all motivate the Ackermann function without this "false" and "harmful" heuristic? – user11235 – 2011-04-18T21:03:27.633

2When I taught elementary teachers the course on arithmetic, they all had been taught that multiplication is repeated addition, but I myself thought it was the cardinality of the cartesian product. We enjoyed discussing this difference in point of view. – roy smith – 2011-05-09T03:06:09.080

1The "repeated addition" characterization has an advantage over the "cardinality of the Cartesian product" characterization (which possibly in some contexts could be considered a disadvantage). That is that it's not self-evident that it's commutative, and so one has a useful exercise for certain kinds of students: figure out why it's commutative. – Michael Hardy – 2011-05-20T02:28:26.707

An early source of multiplication as (area of) a Cartesian product, is Euclid's Elements, where he treats multiplication of two segments as the rectangle with those segments as sides. In Book 2 he proves the usual properties, associativity etc, and even solves the quadratic equation x^2 +cx = c^2 in geometric form. He also shows how to produce a rectangle equal to a given one, but with a given side, thus expressing the product of two segments as a segment, given a unit segment. His proof of III.35 implies if 2 triangles have the same angles their sides have the same geometric cross product. – roy smith – 2011-06-18T20:25:51.083

i meant distributivity, not associativity. – roy smith – 2011-06-20T01:50:23.777


I once very briefly thought that:

Given a vector space $V$ and a sub-space $U \subset V$ that $V-U$ is also a subspace.

I've heard this several times as a TA also.


Posted 2010-05-04T21:02:58.510

Reputation: 940

Why the downvote! I heard this from more than one student in introductory linear algebra classes and when marking. – Benjamin – 2015-05-12T22:21:15.557

5I think this falls under $(x+y)^2=x^2+y^2$, – Thomas Rot – 2015-08-10T12:48:26.577

I never said it always fails, just that it's not generally true and I thought it was for about 1 min once. – Benjamin – 2015-08-10T19:12:30.130

3It always fails... But I don't think this is a common held belief. – Thomas Rot – 2015-08-10T21:40:40.380

@ThomasRot But it always fails, while $(x+y)^2=x^2+y^2$ sometimes holds, especially in characteristic 2. – ACL – 2016-04-21T06:37:34.520

@ACL: What do you mean with but? I don't really thing I understand what you mean here. – Thomas Rot – 2016-04-21T08:18:16.907

1I meant that $V-U$ cannot be a subspace since it doesn't contain 0. On the other hand, in any commutative ring where $1+1=0$, then the formula $(x+ y )^2=x^2+y^2$ holds. – ACL – 2016-04-21T10:02:56.507


I'm not sure how common it is but I've certainly been able to trick a few people into answering the following question wrong:

Given $n$ identical and independently distributed random variables, $X_k$, what is the limiting distribution of their sum, $S_n = \sum_{k=0}^{n-1} X_k $, as $n \to \infty$?

Most (?) people's answer is the Normal distribution when in actuality the sum is drawn from a Levy-stable distribution. I've cheated a little by making some extra assumptions on the random variables but I think the question is still valid.


Posted 2010-05-04T21:02:58.510

Reputation: 283

I don't understand your third paragraph. Are you saying that under the assumptions in the 2nd paragraph, the limiting distribution (rescaling if necessary) is always Levy-stable? – Yemon Choi – 2011-04-12T01:28:41.590

@Yemon, Yes, this is what I was implying. Perhaps I was a little too cavalier? Certainly the sum of (well enough behaved) i.i.d. r.v.'s with power law tails converge to a Levy-Stable distribution... – dorkusmonkey – 2011-04-12T23:53:41.227

4Generally such a limiting distribution doesn't exist. Perhaps you need to divide your sum by the square root of $n$? – John Bentin – 2011-12-29T13:56:17.580


Related to this answer: $$ \pi=\left(\frac{1}{10^5}\sum_{-\infty}^{+\infty}e^{-n^2/10^{10}}\right)^2. $$ Proof: With a computer one can verify that the first 42 billions digits of the two numbers are the same, see J. Borwein and P. Borwein, Strange series and high precision fraud, in The American Mathematical Monthly, 1992, pages 622-640.


Posted 2010-05-04T21:02:58.510

Reputation: 1 955

14I voted this down because I don't think it's a statement that anyone actually believes, and therefore doesn't fit the spirit of this questions, but I have to say it's pretty clever. – Nate Eldredge – 2010-10-19T21:11:58.870

3I would vote it down if I could.

There is nothing false with believing in that Riemann sums converge to the true value of an integral (Poisson in this case). – zhoraster – 2010-10-20T19:12:03.783

10I must admit I'm a little bit surprised just how quickly $f(a) = (1/a \sum e^{-n^2/a^2})^2$ converges to $\pi$ as $a \to \infty$. (According to the identity given in the article, $\lim a^{-2} \log (f(a)-\pi) = -\pi^2$. This feels much faster than we have any right to expect. – Michael Lugo – 2010-10-26T04:40:08.220


I don't know if this is what you are looking for, but I keep hearing that "a differentiable function is one that is locally linear", not one whose local variation can be approximated linearly. No one stops to think about e.g, $x^2$, and the fact that its graph does not look like a line at any value of $x$.


Posted 2010-05-04T21:02:58.510

Reputation: 161

it is also a comment on the imprecision of the words locally, infinitesimally,.... This once led Oort-Steenbrink to give some careful restatements of results previously called as "local Torelli theorems"... – roy smith – 2011-04-14T19:02:56.813


I would say this is more a heuristic than a false statement; as such, it would be more appropriate as an answer to http://mathoverflow.net/questions/2358/most-harmful-heuristic (although I do not think anyone interprets it the way you apparently do).

– Qiaochu Yuan – 2010-05-05T04:53:58.677

Yes, I did not read the question very carefully. I realize it is not a good comment, and, yes, it is more of a abd heuristic than anything else. – Herb – 2010-05-25T23:59:00.800


I had the false belief that recursive functions are always decidable in ZFC.


Posted 2010-05-04T21:02:58.510

Reputation: 317


When I was a kid (8th grade), I solved a bunch of math problems in an exam using the ``well-known identity'' that $(x+y)^2=x^2+y^2$, which I was sure I had been taught the year before. It was of course way before I heard about characteristic two and I didn't get a good grade that day!


Posted 2010-05-04T21:02:58.510

Reputation: 10 318

16Quoth the question, "The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed)". – JBL – 2010-12-01T23:39:33.950

3Also, this is of course just a special case of the more general “law of universal linearity”, which iirc was mentioned in earlier answers… – Peter LeFanu Lumsdaine – 2010-12-02T00:40:49.763


I had a false belief in linear algebra, that a basis of a vector space could have infinitely many elements (like an orthonormal basis in Fourier analysis). That tripped me up trying to understand the definition of tensor products, and even after someone explained the issue to me I didn't believe it at first.


Posted 2010-05-04T21:02:58.510

Reputation: 1

+1 for Gerald's corrected version. – darij grinberg – 2011-03-06T00:16:00.197

7I don't understand. A basis can have infinitely many elements. That's no false belief, that's correct. – Johannes Hahn – 2010-08-22T12:07:02.420

27The false believe would be that when you define basis, you allow infinite linear combinations. If some confusion is possible, say "Hamel basis" ... Even if there is no topology defined, it still will emphasize that only finite linear combinations are considered. – Gerald Edgar – 2010-08-22T12:30:38.497