What are your favorite instructional counterexamples?

147

201

Related: question #879, Most interesting mathematics mistake. But the intent of this question is more pedagogical.

In many branches of mathematics, it seems to me that a good counterexample can be worth just as much as a good theorem or lemma. The only branch where I think this is explicitly recognized in the literature is topology, where for example Munkres is careful to point out and discuss his favorite counterexamples in his book, and Counterexamples in Topology is quite famous. The art of coming up with counterexamples, especially minimal counterexamples, is in my mind an important one to cultivate, and perhaps it is not emphasized enough these days.

So: what are your favorite examples of counterexamples that really illuminate something about some aspect of a subject?

Bonus points if the counterexample is minimal in some sense, bonus points if you can make this sense rigorous, and extra bonus points if the counterexample was important enough to impact yours or someone else's research, especially if it was simple enough to present in an undergraduate textbook.

As usual, please limit yourself to one counterexample per answer.

Qiaochu Yuan

Posted 2010-03-02T05:57:45.173

Reputation: 73 568

3From a pedagogical standpoint, sometimes the minimal counterexample isn't the best one; in particular if it is "too small" to exhibit important general features of what's going on. – benblumsmith – 2011-07-22T12:55:29.060

Someone who knows more about the work being done with the Hasse Principle than I do could probably say something about Selmer's counterexample to an extension of the Hasse-Minkowski theorem to cubic forms. – Ben Linowitz – 2010-03-02T07:04:08.477

@Ben Linowitz. It is in the small and delightful book, J. W. S. Cassels, Lectures on elliptic curves. – Regenbogen – 2010-03-02T14:39:14.490

2@Regenbogen - I am familiar with the proof that Selmer's curve has points everywhere locally but not globally. But that counterexample led many people to study the manner in which the Hasse Prinicple could fail. For example, there is the Brauer-Manin Obstruction. However Skorobogatov has found examples of curves with trivial Manin obstruction and everywhere local points but no global points, so the story is not finished...In my comment I was suggesting that someone more familiar with the current work might use this example. – Ben Linowitz – 2010-03-02T17:23:02.750

1

@Ben Linowitz. Oh I am sorry for saying irrelevant things. I must confess I do not know anything at all. Maybe the following MSRI video might interest you(if you were not already aware of it)... http://www.msri.org/communications/vmath/VMathVideos/VideoInfo/3821/show_video

– Regenbogen – 2010-03-02T17:38:33.617

Answers

111

The matrix $\left(\begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix}\right)$ has the following wonderful properties. (Feel free to add or edit; I can't remember all the reason I loathed it when I was learning linear algebra. It's funny how unexciting they all now seem, but it's a counterexample for almost every wrong linear algebra proof I tried to give.)

  • Only zeroes as eigenvalues, but non-zero minimal polynomial (in particular, the minimal polynomial has bigger degree than the number of eigenvalues). Probably my favorite way to state this fact: the minimal polynomial is not irreducible or square-free. The same thing in a fancier language: the Jordan canonical form is not diagonal.

  • Not diagonalizable, even over an algebraically closed field.

  • Not divisible over $\mathbb C$. There are no matrices $M$ and integers $n\ge2$ so that $M^n = \left(\begin{smallmatrix}0 & 1\\\ 0 & 0\end{smallmatrix}\right).$ All diagonalizable and most non-diagonalizable complex matrices have $n$th roots.

    (This is because, if there was a square root, it'd have minimal polynomial x4, but since it's a two-by-two matrix, Cayley-Hamilton implies that the characteristic polynomial has degree 2).

  • The matrix is nilpotent but not zero.

  • It's one of the best examples when you need to remember why matrix multiplication is not commutative.

  • Thinking of k2 as a k[x]-module where x acts as this matrix should give wonderful (counter)-examples of modules for all the same reasons.

Also, $\left(\begin{smallmatrix}1 & 1\\ 0 & 1\end{smallmatrix}\right)$ is an example of an invertible matrix with the first three properties above. Its action on k2 is in some sense the simplest example of a representation of a group ($\mathbb{Z}$) which is indecomposable but not irreducible.

Ilya Grigoriev

Posted 2010-03-02T05:57:45.173

Reputation: 2 158

2You mean that most non-diagonalizable square matrices have $n$th roots. – Kevin O'Bryant – 2011-03-27T23:50:13.570

2"the minimal polynomial is not irreducible"

I am not sure I get your point. Minimal polynomials are usually not irreducible. Did you mean "does not have to be multiplicity-free"? – Vladimir Dotsenko – 2010-03-03T14:38:24.483

@Vladimir: Yes and no. In other context (Galois theory) minimal polynomials are always irreducible. But you're right, I was a bit careless; I've edited it now. – Ilya Grigoriev – 2010-03-04T00:07:01.697

13I think it's better to remember that functions are not commutative: if f puts pants on a person and g puts underpants on them, then f(g(naked person)=half-naked person, and g(f(naked person)=half-naked superman. – Michał Masny – 2015-03-08T15:23:43.103

Of course $\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$ doesn't have the first of the three properties (only zero as an eigenvalue)! – LSpice – 2016-05-13T02:35:22.430

I don't follow your comment that the invertible matrix has only zero eigenvalues (one of the first three properties you list). Isn't that impossible? – Nick Salter – 2010-07-29T16:31:48.657

@Nick: You are right. I just mean that the minimal polynomial is not irreducible. – Ilya Grigoriev – 2010-07-29T17:44:56.813

106

The Fabius function, everywhere $C^\infty$, nowhere analytic.

image

see... sci.math post

references:
J. Fabius, "A probabilistic example of a nowhere analytic $C^\infty$-function". Z. Wahrsch. Verw. Geb. 5 (1966) 173--174.

K. Stromberg, PROBABILITY FOR ANALYSTS (Chapman & Hall, 1994), pp. 117--120.

Gerald Edgar

Posted 2010-03-02T05:57:45.173

Reputation: 25 496

4

Another link (math forum): http://mathforum.org/kb/message.jspa?messageID=508877&tstart=0

– Gerald Edgar – 2012-11-04T13:56:41.007

Good one,Gerald-whoa! – The Mathemagician – 2010-04-04T19:20:05.923

23Very cool! I'm going to print out the definition and bring it to every physics class I'm in from now on. :) – Vectornaut – 2010-04-04T22:02:49.930

2

a good explanation can also be found here: http://calculus7.org/tag/probability

– Manfred Weis – 2015-08-08T03:01:29.093

I was just thinking "I wonder if there any pathological examples of a smooth function." Now I know. – PyRulez – 2015-09-08T00:37:36.000

4The link has gone bad (takes you to "google groups"). – Victor Protsak – 2010-09-30T04:01:18.277

85

A polynomial $p(x) \in \mathbb{Z}[x]$ is irreducible if it is irreducible $\bmod l$ for some prime $l$. This is an important and useful enough sufficient criterion for irreducibility that one might wonder whether it is necessary: in other words, if $p(x)$ is irreducible, is it necessarily irreducible $\bmod l$ for some prime $l$?

The answer is no. For example, the polynomial $p(x) = x^4 + 16$ is irreducible in $\mathbb{Z}[x]$, but reducible $\bmod l$ for every prime $l$. This is because for every odd prime $l$, one of $2, -2, -1$ is a quadratic residue. In the first case, $p(x) = (x^2 + 2 \sqrt{2} x + 4)(x^2 - 2 \sqrt{2} x + 4)$. In the second case, $p(x) = (x^2 + 2 \sqrt{-2} x - 4)(x^2 - 2 \sqrt{-2} x - 4)$. In the third case, $p(x) = (x^2 + 4i)(x^2 - 4i)$. This result can be thought of as a failure of a local-global principle, and the counterexample is minimal in the sense that the answer is yes for quadratic and cubic polynomials.

Qiaochu Yuan

Posted 2010-03-02T05:57:45.173

Reputation: 73 568

13Similarly, if the units modulo n are not cyclic, then the nth cyclotomic polynomial \Phi_n(x) will be reducible mod p for all p. – Ben Linowitz – 2010-03-02T06:15:50.747

51Even better, the polynomial $x^4-72x^2+4$ is irreducible in $\mathbb{Z}[x]$, but reducible modulo every <I>integer</I>. (Dummit and Foote, 3rd edition, page 309) – Alfonso Gracia-Saz – 2010-03-02T16:41:37.843

22The polynomial $(x^2+31)(x^3+x+1)$ has a root modulo every prime but no roots in Q. No polynomial of degree < 5 has this property. – AVS – 2010-04-04T20:21:50.737

Good example! I think it's also important to note that no irreducible polynomials have this property by the Frobenius density theorem. – Qiaochu Yuan – 2010-04-04T21:32:13.177

1@Qiaochu Yuan: Yes. Here one also uses the fact that every transitive permutation group contains an element that fixes no points. – AVS – 2010-04-05T11:40:48.253

78

I like the double sequence $a_{nm} = \frac{n}{n+m}$ to show that $\lim_{n\to\infty}\lim_{m\to\infty} a_{nm}\neq \lim_{m\to\infty}\lim_{n\to\infty} a_{nm}$ .

Dirk

Posted 2010-03-02T05:57:45.173

Reputation: 6 399

14

The sequence which is 0 or 1 depending on which of m and n is greater also works. See the fifth example from http://www.tricki.org/article/Just-do-it_proofs and its accompanying discussion.

– aorq – 2011-03-19T18:18:16.943

26Yeah, that works, too! However, I experienced that undergraduates sometimes feel a bit uncomfortable with such "piecewise" definitions and are more happy with a more "natural" example (whatever that means). – Dirk – 2011-03-19T19:58:18.090

2@aorq Iverson bracket: $[m>n]$ – Akiva Weinberger – 2015-09-02T00:39:34.463

Would $a_{nm}=\frac{n}{m}$ also be such a sequence? – Ayush Khaitan – 2017-01-23T20:29:18.440

3In principle, yes, but then you also have "improper limits" (i.e. convergence to plus infinity), and I in Germany we teach these separately. – Dirk – 2017-01-23T21:31:50.510

72

The Cantor set is a nice source of counterexamples:

The first measure zero sets you meet are usually countable. However, the Cantor set is uncountable and measure zero.

It is totally disconnected, yet it is not a discrete space. In particular, this shows that connected components of a topological space need not be open sets.

Dinakar Muthiah

Posted 2010-03-02T05:57:45.173

Reputation: 2 918

6This is a very good answer. I would go so far as to say that if you're studying general topology but haven't encountered the Cantor set, your ideas of what a topological space can be are fundamentally incomplete. – Pete L. Clark – 2010-12-29T06:49:38.357

7“It is totally disconnected, yet it is not a discrete space.” As a professional matter, I like $\mathbb Q_p$ even better as an example of this behaviour. (Of course, topologically it's nearly the same!) If I may piggy-back on Pete's comment, if you haven't encountered $\mathbb Q_p$, then your ideas of what a complete metric space can be are fundamentally incomplete. :-) – LSpice – 2011-03-28T16:03:31.757

6also positive measure cantor set is a very nice example to difference betwean meagre and null sets – Ostap Chervak – 2011-04-10T09:57:45.177

1Take its union with the set of midpoints of the Cantor set's "gaps" (gap = connected component of Cantor set's complement in $[0,1]$), to get an interesting space: An uncountable set with a countable, dense and isolated subset! (The subset being the set of gap-midpoints.) // I found this after a class on the topology on the ultrafilters on $\Bbb N$. A homework problem was to show that the principle ultrafilters were dense in the space, but each was isolated. It continued, "Try to draw this, and become sad." So I was surprised to find out that I actually could draw this situation. – Akiva Weinberger – 2015-09-02T00:38:12.653

The Cantor set is also a complete metric space, in a few interesting ways. – Toby Bartels – 2017-02-09T21:24:30.127

65

A counter-example in graph theory - the Petersen graph.

alt text http://www.imada.sdu.dk/~btoft/GT2009/PetersenGraphEmbeddings_800.gif

In many ways it is the most simple graph with many strange properties. See the article on Wiki.

Quote from our professor who teaches graph theory:

If you think you've proved any lemma about graphs, try Petersen first!

Hsien-Chih Chang 張顯之

Posted 2010-03-02T05:57:45.173

Reputation: 800

57

The 8-element quaternion group. It can't be reconstructed from its character table (D_4 has the same one), and every subgroup is normal but it's not abelian.

Allen Knutson

Posted 2010-03-02T05:57:45.173

Reputation: 20 250

+1, darij: this is perhaps worthy of its own answer. – Pete L. Clark – 2010-12-29T06:52:43.797

1Also it has nontrivial inner and outer automorphisms, and (for minimality) is the smallest group with this property. This makes it a good example for distinguishing conjugate subgroups from the broader class of subgroups that are carried to each other by an automorphism. The klein 4-group can also be used for this, but the quaternion group is more pedagogically satisfying because it has a nontrivial characteristic subgroup as well as the 3 that are normal but not characteristic. (It is minimal for this property as well.) – benblumsmith – 2011-07-22T12:52:35.180

23And it has quaternionic representations - something that serves as counterexample to many beginner's conjectures in representation theory (a la "any representation can be constructed in the smallest field where its character lives"). – darij grinberg – 2010-04-04T15:59:44.277

52

I'm surprised no one mentioned the Hawaiian Earring:

alt text

It's path-connected but not semi-locally simply connected (because any small neighborhood of the origin must contain a non-contractible loop). This implies many interesting properties, which make it a great counter-example. For instance...

  • The Hawaiian Earring cannot have a universal cover.
  • The Hawaiian Earring is not a CW-complex, although it is a compact, complete metric space
  • An example of a space which is semi-locally simply connected and simply connected but is not locally simply connected is the cone on the Hawaiian Earring.
  • For many years people thought the fundamental group was always a topological group. This turns out to be false, thanks to the Hawaiian Earring. There's a nice post about this here on MO
  • This question is Community Wiki for a reason. I'm sure there are other examples of conjectures the Hawaiian Earring has disproven, so please add them!

David White

Posted 2010-03-02T05:57:45.173

Reputation: 9 469

3I like the fact, that it's fundamental group is uncountable. This is a vivid example for showing students, which are new to algebraic topology, that the fundamental group are not just "some" loops in the space. – archipelago – 2013-03-23T13:39:49.557

8The analogs in higher dimensions have nonzero homology in arbitrarily high dimensions! – Jeff Strom – 2013-03-23T14:48:49.523

46

The Weierstrass function - which I guess is a counterexample to the conjecture that a function which is continuous everywhere must be differentiable somewhere. I remember being pretty amazed when I first encountered it. It made me realize that continuity and differentiability are really different notions.

Tony Huynh

Posted 2010-03-02T05:57:45.173

Reputation: 18 841

I was really glad when my analysis professor first showed this example; I had the same realization that you did. – Cory Knapp – 2010-03-04T03:15:24.907

I think most of us do.I actually was shown several variamts,including van der Wearden's. A very good thing indeed to show newbies in real analysis. – The Mathemagician – 2010-04-04T19:21:32.590

42

The basic fact that there are smooth non-analytic functions on $\mathbb R$, and that there are compactly supported smooth functions, is important in real analysis and functional analysis.

$f(x) =\begin{cases} \exp(-1/(1-x^2)),& x \in (-1,1) \\\ 0& \text {otherwise} \end{cases}$

alt text

The usual examples of these functions often seem contrived. Here are examples of smooth nowhere analytic functions.

Douglas Zare

Posted 2010-03-02T05:57:45.173

Reputation: 24 902

Hey, more cool information on smooth, nowhere-analytic functions! – Vectornaut – 2010-04-04T22:10:18.957

How is this compactly supported? I thought the support is where it's nonzero, and this is nonzero on $(-1,1)$ which isn't compact. – Akiva Weinberger – 2015-09-02T01:16:46.197

@columbus8myhw: The support is the closure of where it is nonzero. https://en.wikipedia.org/wiki/Support_(mathematics)

– Douglas Zare – 2015-09-03T07:09:11.193

@columbus8myhw The support is the closure of the set where the function is not zero. – Dirk – 2015-09-03T18:15:45.360

Ah, that makes sense. – Akiva Weinberger – 2015-09-03T18:16:36.140

36

Counterexamples are very important when a student learns how to think in intuitionistic logic (and he has already been "spoiled" by classical logic). The counterexamples destroy the classical intuition, and when properly explained they help the student understand how to think intuitionistically. Some that seem to work praticularly well in my experience involve finite sets. Intuitionistically the following are not provable:

  1. A subset of a finite set is finite.
  2. The powerset of a finite set is finite.
  3. If a subset of $\mathbb{N}$ is not finite then it is infinite.
  4. The elements of a finite set may be listed without repetition.

All of these can be rescued with the additional assumption that the sets involved have decidable equality and that the subsets involved have decidable membership.

However, it does not really help the student to just know that certain "obvious" facts are not provable. He really needs to see how the "facts" can be false. The ones listed above are all false in the effective topos, but that's a complicated gadget for a beginner. It turns out informal explanations work well enough because most students know a little bit of programming. They just needs to know that the Halting Oracle does not exist.

My favorite counterexample in intuitionistic logic is that it is consistent to assume the so-called Axiom of Enumerability, which says that there are countably many countable subsets of $\mathbb{N}$. (Explanation: in the effective topos this just means that there is an effective enumeration of computably enumerable subsets of $\mathbb{N}$.) Many basic theorems of computability theory can be proved, phrased in a suitable form, from the axiom of enumerability using just constructive logic and no mention of machines of any kind.

Andrej Bauer

Posted 2010-03-02T05:57:45.173

Reputation: 26 815

4Which definitions of "finite set" and "infinite set" are you using? – aorq – 2010-03-02T14:33:50.430

3A set $A$ is finite if there exists a natural number $n$ and an onto map $e : \lbrace{1, ..., n\rbrace} \to A$. A set $B$ is infinite if there exists 1-1 map $m : \mathbb{N} \to B$, i.e., $B$ contains an infinite sequence without repetitions. – Andrej Bauer – 2010-03-02T21:34:23.823

Thanks, Andrej; These are all really fun... now to pick them apart and internalize them. – Cory Knapp – 2010-03-04T03:11:10.173

36

I've always been fond of the popcorn function (aka Thomae's Function), which is given by $f\colon \mathbb{R} \to \mathbb{R}$ via

$f(x) = \begin{cases} \frac{1}{n} & \mbox{if } x = \frac{m}{n} \in \mathbb{Q} \\ 0 & \mbox{if } x \notin \mathbb{Q}. \end{cases}$

This function has a couple of amusing properties.

(1) It is upper semicontinuous on $\mathbb{R}$, yet has a dense set of discontinuities (every one of which is removable) (namely $\mathbb{Q})$.

(2) Since it is bounded and has a set of measure zero as its set of discontinuities, it is Riemann integrable. So if we consider $g(x) = \int_0^x f(t)\ dt$, we see that $g \equiv 0$, so that $g'(x) \not \hskip 2pt = f(x)$ on a dense set.

References: http://en.wikipedia.org/wiki/Thomae%27s_function and of course "Counterexamples in Analysis" (Sec 2.15-2.17)

Jesse Madnick

Posted 2010-03-02T05:57:45.173

Reputation: 371

4John H. Conway called that the "Stars over Babylon" function in a class I took. Finally I know another name. – Elizabeth S. Q. Goodman – 2011-03-17T07:35:36.447

And its "reciprocal" ($g(x)=\begin{cases}n,&x=m/n\0,&x\notin\Bbb Q\end{cases}$) is also interesting, in that it's unbounded on every interval. Not quite as weird as Conway's base 13 function, which has an image of $\Bbb R$ on every interval, but still a nice little pathology.

– Akiva Weinberger – 2015-09-02T01:27:13.850

32

The Baumslag--Solitar groups have presentations

$BS(p,q)=\langle a,b\mid a^p=b^{-1}a^q b\rangle$.

They have the following nice properties:

  1. they're two generator, one relator groups;
  2. they can be written as an HNN extension of $\mathbb{Z}$ over $\mathbb{Z}$. (This means that they're constructed by 'gluing' $\mathbb{Z}$ to itself in some way.)

So from the point of view of combinatorial group theory, they could hardly be simpler. And yet, for suitable values of $p$ and $q$ (typically $p,q$ relatively prime integers greater than 1 will do), we find that:

  1. they're non-Hopfian, meaning that they admit a self-epimorphism with non-trivial kernel;
  2. hence they're not even residually finite;
  3. they have exponential Dehn function (meaning that the word problem can be solved, but only very slowly);
  4. their virtual first Betti number is one (meaning that every finite-index subgroup has abelianisation of rank one)...

I could go on.

HJRW

Posted 2010-03-02T05:57:45.173

Reputation: 17 048

2

And for p != q they can't be the fundamental group of a 3-manifold -- see http://mathoverflow.net/questions/6132/fundamental-group-of-3-manifold-with-boundary for some references.

– Steven Sivek – 2010-03-02T20:32:07.893

That the Dehn function is big does not mean that the word problem can be solved only slowly. – YCor – 2015-08-22T12:20:47.630

@YCor - of course that is strictly correct. I was trying to give some rough idea of what it meant. – HJRW – 2015-08-28T21:38:02.723

31

The matrices $A=\begin{pmatrix} 17\times 11 + 1 & 25\times 11\\ 11^2 & 16\times 11 + 1 \end{pmatrix}$ and $B = \begin{pmatrix} 17\times 11 + 1 & 11 \\ 25\times 11^2 & 16\times 11 + 1 \end{pmatrix}$ are similar modulo $m$ for every positive integer $m$ but are not similar over the integers.

In other words, there exist matrices $X_m\in GL_2(\mathbf Z/m\mathbf Z)$ such that $X_mA \equiv BX_m \mod m$ for every $m$, but there does not exist a matrix $X\in GL_n(\mathbf Z)$ such that $XA = BX$.

This is due to Stebe, Conjugacy separability of groups of integer matrices. Proc. Amer. Math. Soc., 32:1–7, 1972.

Amritanshu Prasad

Posted 2010-03-02T05:57:45.173

Reputation: 3 675

30

Let $P dx + Q dy$ be a one-form, or if you're using the terminology of an introductory multivariable calculus course, a "vector field" that you can take line integrals of. Then students learn Green's Theorem, which says that if some countour $C$ bounds a region $D$, then $$\int_C P dx + Q dy = \int_D \left(\frac{dQ}{dx}-\frac{dP}{dy}\right) dx dy.$$

From this, one deduces that if the expression on the right hand side vanishes, then the integral around any contour is $0$. In particular, this allows one to define a primitive for $P dx + Q dy$.

Many students (myself included, a long way back) don't pay enough attention to the hypotheses in Green's Theorem and then assume that this is true of the following one-form (or "vector field"), which is my fundamental counterexample:

$$\frac{-ydx}{x^2+y^2} + \frac{xdy}{x^2+y^2}$$

Eventually a student discovers that the integral of this around the origin is $2 \pi$ and then wonders what went wrong. The problem is that the hypothesis of Green's Theorem requires that the form be defined everywhere in $D$.

In other words, this is a fundamental counterexample to the claim that a one-form in the plane with zero curl (where by "curl" I just mean the right hand side of the above) has a primitive.

Furthermore, this is a fundamental example of a nontrivial element in a de Rham cohomology group. In this case, the one-form above generates $H^1_{\text{dR}}(\mathbb{R}^2\setminus \{(0,0)\},\mathbb{R})$.

David Corwin

Posted 2010-03-02T05:57:45.173

Reputation: 6 150

27

The Poincaré homology sphere, a spherical 3-manifold with fundamental group the binary isosahedral group, was Poincaré's counterexample to the original formulation (in terms of homology) of his conjecture. Due to its countless descriptions -- as a spherical 3-manifold, via Dehn surgery, as the configuration space of an isosahedron, etc -- it's still a motivational example in geometry and topology.

Matthew Stover

Posted 2010-03-02T05:57:45.173

Reputation: 511

26

Added in edit. To understand this answer, one needs a definition that is current in French but for which I have not found a clearly equivalent word in English (causing misunderstandings, as can be seen in the comments). One says that a function $f$ defined near $0$ (say) has a "développement limité" (DL, which could be translated by "restricted expansion") of order $n$ at $0$ if there are numbers $a_0,a_1,\dots, a_n$ such that

$$ f(x) = a_0+a_1 x+a_2 x^2 + \dots + a_n x^n + o_{x\to 0}(x^n). $$

My instructional counter-example is: (end of addition)

The function $x\mapsto x^3\sin(1/x)$ has a second-order DL (before edit: "Taylor series") but is not twice differentiable at $0$. The circumstances where I came across this example are too embarrassing to tell here...

Benoît Kloeckner

Posted 2010-03-02T05:57:45.173

Reputation: 10 345

1What is the second-order Taylor series for this function? – Mehrdad – 2015-03-30T08:15:13.883

@Mehrdad: since $\sin(1/x)=0(1)$, we have $x^3\sin(1/x)=0+O(3)$: the second-order Taylor series is trivial. – Benoît Kloeckner – 2015-03-30T16:54:57.433

But I don't understand, isn't the 3rd term in the 2nd order Taylor series based on the 2nd derivative of the function? How can you say it has a Taylor series when the definition requires its derivatives? – Mehrdad – 2015-03-30T17:11:53.097

6@Mehrdad That's precisely my point: the Taylor-Peano formula deduces a Taylor series from the derivatives, but a Taylor series may exist even when the higher-order derivative don't. – Benoît Kloeckner – 2015-03-30T18:50:48.777

(In my first comment $O(3)$ should of course be $O(x^3)$) – Benoît Kloeckner – 2015-03-30T18:52:41.630

I mean I don't understand how you can define a Taylor series when the function doesn't have derivatives. That's why I asked how you found the series, because it couldn't have been by taking derivatives... – Mehrdad – 2015-03-30T19:35:12.193

Let us continue this discussion in chat.

– Benoît Kloeckner – 2015-03-30T20:14:55.490

5Somehow it seems less snappy when you phrase this as saying that $x \mapsto x\sin(1/x)$ has a $0$th-order Taylor series at $0$ …. – LSpice – 2016-05-13T04:41:53.643

1Would you also say that the function $f(x) = x^2 \cdot I{\mathbb{Q}}(x)$ (where $I\mathbb{Q}$ is the indicator function of the rationals) has a second order Taylor expansion at $x=0$? – Steven Gubkin – 2018-01-11T19:17:46.303

@StevenGubkin: I would. Note that in French, we have much different words: "série de Taylor" for the series obtained using derivatives, "développement limité" for an expression of the form polynomial + landau-notation-remainder. – Benoît Kloeckner – 2018-01-11T21:19:01.670

1I think that by "second-order Taylor series" you actually mean "second-order Taylor polynomial". And as other people are getting at in the comments, whether or not your claim is true depends on your exact definition of the term "Taylor polynomial". – tparker – 2018-02-05T03:44:42.763

@tparker: in fact it seems there is no real equivalent to the French word (e.g. Wikipedia does not match the French article to any article in any other language). I will make my answer more explicit. – Benoît Kloeckner – 2018-02-05T14:32:31.233

@LSpice: long overdue answer to your comment: the point is that we really have a second order series/expansion, because the remainder is $o(x^2)$, not $o(1)$. – Benoît Kloeckner – 2018-02-05T14:41:24.703

1@BenoîtKloeckner, thanks for the response! I agree with you that $x \mapsto x^3\sin(1/x)$ has such a remainder, but I was (in a feeble attempt at witticism) talking about $x \mapsto x\sin(1/x)$, for which the remainder is $\mathrm o(1)$, not $\mathrm o(x^2)$. – LSpice – 2018-02-05T16:12:04.940

@LSpice Oh, sure, I read too fast! – Benoît Kloeckner – 2018-02-05T17:26:46.773

2I flunked an exam in ordinary differential equations getting that one wrong-trust me,the circumstances can't be worse then that. – The Mathemagician – 2010-07-29T18:28:23.267

26

A basic result in commutative algebra asserts that direct limits commute with tensor products. My favourite counterexample to the statement obtained by replacing "direct" with "inverse" is the following. Let $p$ be a prime number; then

$\bigl(\varprojlim_n\mathbb Z/p^n\mathbb Z\bigr)\otimes_{\mathbb Z}\mathbb Q\cong\mathbb Q_p$,

the field of $p$-adic numbers (completion of $\mathbb Q$ with respect to the metric induced by the $p$-adic valuation), while

$\varprojlim_n\bigl((\mathbb Z/p^n\mathbb Z)\otimes_{\mathbb Z}\mathbb Q\bigr)=0$,

since every $\mathbb Z/p^n\mathbb Z$ is torsion and $\mathbb Q$ is divisible.

Stefano V.

Posted 2010-03-02T05:57:45.173

Reputation: 386

24

The Moulton plane is a projective plane that is a counterexample to the Desargues theorem, the little Desargues theorem, and just about every "nice" property of projective planes.

Its discoverer, F.R. Moulton, is best known as an astronomer. He apparently came up with the Moulton plane after sitting in on a projective geometry course as a graduate student.

John Stillwell

Posted 2010-03-02T05:57:45.173

Reputation: 9 944

1Part of why Desargues "Theorem" is intriguing is that it holds in some projective planes and not in others. There are finite planes where it holds and finite planes where it does not hold. If there is a way to introduce coordinates for the plane with numbers from a division ring then then Desargues Theorem will hold. It also holds for projective planes sitting in higher dimensional projective spaces. In the real projective plane the theorem holds. The Moulton plane is a fascinating example. – Joseph Malkevitch – 2010-03-03T14:05:57.033

24

Homotopy groups do not, in general, commute with sequential colimits, even for nice maps between nice spaces.

I just learned this beautiful example from Bill Dwyer.
Take the sequence

$S^1\stackrel{2}{\longrightarrow}S^1\stackrel{3}{\longrightarrow}S^1\stackrel{4}{\longrightarrow}\cdots.$

Here $n$ denotes the $n$th power map on $S^1$. Thinking of $S^1$ as $\mathbb{R}/\mathbb{Z}$, one finds that the colimit of this sequence (in the category of topological spaces) is the quotient group $\mathbb{R}/\mathbb{Q}$. Note that this quotient group, topologized as a quotient space of $\mathbb{R}$ by the relation $x\sim y$ if $x-y\in \mathbb{Q}$, has the indiscrete topology. In particular, the colimit of this sequence is a contractible topological space and has trivial homotopy groups.

On the other hand, the colimit of the corresponding sequence of fundamental groups is the group $\mathbb{Q}$ (checking this is a fun exercise).

(There's something sort of odd here, because one might have guessed that $\mathbb{R}/\mathbb{Q}$ would be a model for $K(\mathbb{Q}, 1)$, since after all $\mathbb{R}$ is a free $\mathbb{Q}$-space. But there are no interesting open sets in the quotient and hence no chance of local triviality.)

Dan Ramras

Posted 2010-03-02T05:57:45.173

Reputation: 5 258

22

The blowup of $\mathbb{P}^2$ in the 9 points of intersection of two generic cubics admits infinitely many $(-1)$ curves. This example is very important in getting rid of the naif picture of algebraic surfaces.

Andrea Ferretti

Posted 2010-03-02T05:57:45.173

Reputation: 8 136

4Even though the idea of "blowing down" is to get rid "a" (-1) curve. You can always blow down finitely many times to get rid of all of them, even if you start infinitely many. This is a nice example of that case. – Matt – 2010-04-04T17:38:20.863

21

I'd say the Tutte Graph, which is a counterexample to

Tait's conjecture: Every 3-connected cubic planar graph has a Hamiltonian cycle.

Initially, I thought this counterexample was extremely non-instructive, since I assumed that Tutte discovered it via some ingenious trial and error. But, after seeing a talk by Bill Cunningham, I discovered how Tutte came up with his counterexample and why it is a counterexample (it's unclear from looking at Tutte's graph that it is not Hamiltonian). The idea is quite simple but useful. Tutte assumed that Tait's conjecture was true and proceeded to prove a sequence of stronger (yet equivalent) conjectures. He then found a very small counterexample to the strongest conjecture, and then deconstructed the sequence of proofs to obtain a counterexample to Tait's conjecture.

I really like this method because it shows that there is hope for a mathematical caveman like me as long as I use my brain. That is, the counterexample was actually not pulled out of thin air like I initially thought.

Tony Huynh

Posted 2010-03-02T05:57:45.173

Reputation: 18 841

21Very few results are pulled out of thin air. The problem is, mathematical culture has not favored showing where they do come from. – Mariano Suárez-Álvarez – 2010-03-03T16:56:37.183

8I'd add that one of the reason for this is that journals have a strong preference for short articles, although today, with internet distribution, I do not see any reason for this.

As a result, mathematicians cut out the why from the proofs, leaving only the necessary steps. But this is of course discussed in other threads... – Andrea Ferretti – 2010-03-03T17:09:04.307

21

A standard result in introductory calculus classes is that, if a function has positive derivative on an open interval, then it's increasing there.

Based on this, students tend to think that, if $f'(a)>0$, then $f$ must be increasing "near $a$."

However, the example $f(x) = 2x^2\sin(1/x)+x$ (set $f(0)=0$) shows that this is quite false!

Ramsey

Posted 2010-03-02T05:57:45.173

Reputation: 2 258

3

See On Functions Increasing at a Point at http://clem.mscd.edu/~talmanl/PDFs/APCalculus/IncrAtPt_New.pdf.

– I. J. Kennedy – 2012-09-04T22:36:40.847

20

The alternating group on 4 letters is nice because it provides a counterexample to the converse of Lagrange's theorem: It has order 12, but it does not have a subgroup of order 6.

David

Posted 2010-03-02T05:57:45.173

Reputation: 1

3Similarly, the symmetric group on 5 letters has no subgroup of order 15, since (up to isomorphism) the only group of order 15 is the cyclic group, and $S_5$ has no element of order 15. – Gerry Myerson – 2011-03-27T23:35:48.757

19

Finite topological spaces often provide nice and simple counterexamples in topology, including algebraic topology (check J. Barmak's thesis). After getting familiar with those spaces one easily comes up with examples of phenomena such as weakly homotopy equivalent spaces which are not homotopy equivalent (spaces consisting of 4 points and 6 points suffice) or homomorphisms between homology/homotopy groups that are not induced by continuous maps.

Of course, other counterexamples are available, but finite ones are certainly minimal in a sense.

Michał Kukieła

Posted 2010-03-02T05:57:45.173

Reputation: 1 259

15

The Warsaw circle $W$ http://en.wikipedia.org/wiki/Continuum_%28topology%29 is a counterexample for quite a number of too naive statements.

The Warsaw circle can be defined as the subspace of the plane $R^2$ consisting of the graph of $y = \sin(1/x)$, for $x\in(0,1]$, the segment $[−1,1]$ in the $y$ axis, and an arc connecting $(1,\sin(1))$ and $(0,0)$ (which is otherwise disjoint from the graph and the segment).

Some observations: $W$ is weakly contractible (because a map from a locally path connected space cannot ''go over the bad point'').

Let $I$ denote the segment $[−1,1]$ in the $y$ axis. Then $W/I\cong S^1$ is just the usual circle, and thus we have a natural projection map $g:W \to S^1$. The point-preimages of $g$ are either points or, for a single point on $S^1$, a closed interval.

Thus the assumptions of the Vietoris-Begle mapping theorem hold for $g$, proving that $g$ induces an isomorphism in Cech cohomology. Thus the Cech cohomology of $W$ is that of $S^1$, but it has the singular homology of a point, by Hurewicz.

Since $I\to W$ is an embedding of compact Hausdorff spaces, we have an induced long exact sequence in (reduced) topological $K$-theory (see, for example, Atiyah's $K$-theory Proposition 2.4.4). Since $I$ is contractible, we get that $W$ and $S^1$ also have the same topological $K$-theory.

Note that the Warsaw circle is a compact metrizable space, being a bounded closed subspace of $R^2$. By looking on points on $I$ one sees that $W$ is not locally path-connected (and, in particular, not locally contractible).

The above observations imply:

  1. A map with contractible point-inverses does not need to be a weak homotopy equivalence, even if both, source and target, are compact metric spaces. Assuming that the base and the preimages are finite CW complexes does not help.

  2. The Vietoris-Begle Theorem is false for singular cohomology (in particular, the wikipedia version of that Theorem is not quite correct).

  3. The embedding $I\to W$ cannot be a cofibration in any model structure on $Top$, where the weak equivalences are the weak homotopy equivalences and the interval $I$ is cofibrant. Because then we would have a cofiber sequence $I\to W\to S^1$ and thus also a long exact sequence in singular cohomology.

  4. $W$ does not have the homotopy type of a CW complex (since it is not contractible).

  5. Even though the map $g$ is trivial on fundamental groups, it does not lift to the universal cover $p: \mathbb{R} \to S^1$, because $g$ cannot be nullhomotopic. Thus the assumption of local path connectivity in the lifting theorem is necessary.

Johannes Ebert

Posted 2010-03-02T05:57:45.173

Reputation: 15 823

15

My favourite counterexample is purely academic: it does not have any applications, but I think it is pretty.

Let $X = \mathbb{N} \times \mathbb{N}$. Define a non-empty set $U \subseteq X$ to be open if for cofinitely many $x \in \mathbb{N}$ the set $\{ y \in \mathbb{N} \vert (x,y) \in U\}$ is cofinite.

Construct a sequence in $X$ that hits every point in $X$ exactly once. In other words, take a bijection $\mathbb{N} \rightarrow X$. Then:

  • $X$ is countable;
  • every point in $X$ is an accumulation point of this sequence, but
  • the sequence has no convergent subsequences.

In particular, this is an example in a countable set that accumulation point of a sequence does not have to be a limit of a subsequence. I call this the Herreshoff topology for the (high-school) student of mine who came up with it. (I could not find it anywhere else, although I do not discard that I did not look hard enough.)

Alfonso Gracia-Saz

Posted 2010-03-02T05:57:45.173

Reputation: 247

3

This is sort of Arens-Fort space: http://en.wikipedia.org/wiki/Arens-Fort_space. Here you have the neighbourhoods of (0,0) in that space (considering (0,0) to be not in N x N), where there N x N is discrete. Your sequence then has the same properties.

– Henno Brandsma – 2010-03-06T08:40:50.763

Yes, you are right. Thanks. – Alfonso Gracia-Saz – 2010-03-10T05:28:04.273

14

There exists a $3$-dimensional smooth projective variety $X$ which cannot be birational to a smooth variety with nef canonical bundle. This is because $K_X$ is big; if it was also nef it would have no cohomology and we could compute its self-intersection with Riemann-Roch by looking at the number of sections of its powers. It turns out that the self-intersection would be $3/2$.

This example (by Reid, I think) shows that if you want to have minimal models you have to allow singular varieties, so that $K_X$ can still be defined, but is not a Cartier divisor. This has led to the whole branch of birational geometry studying the type of singularities which are allowed in the minimal model program, like terminal, canonical, log-terminal, KLT and so on.

Andrea Ferretti

Posted 2010-03-02T05:57:45.173

Reputation: 8 136

14

Small's Example from noncommutative algebra...

The triangular ring $T = \pmatrix{\mathbb{Z} & \mathbb{Q} \\\ 0 & \mathbb{Q}}$ has the following properties:

  • It's right noetherian but not left noetherian
  • It's right hereditary but not left hereditary
  • The right global dimension is 1 but the left global dimension is 2
  • This generalizes to give an example of a ring with right global dimension $n$ and left global dimension $n+1$ by replacing $\mathbb{Z}$ by $R$, a commutative noetherian domain of global dimension $n$, then replacing $\mathbb{Q}$ by $K = Frac(R)$
  • A similar example gives a ring which is noetherian but neither left nor right Ore. Just take $R = \pmatrix{S & 0 \\\ S & I}$ where $S = \pmatrix{\mathbb{Z} & 0 \\\ \mathbb{Z}_p & \mathbb{Z}_p}$ and $I = \pmatrix{\mathbb{Z} & 0 \\\ 0 & \mathbb{Z}_p}$ is an $S$-ideal.

Having been trained to think in a commutative world, I found the existence of an example for any one of these to be surprising. The fact that they were all (basically) the same example is even more amazing.

David White

Posted 2010-03-02T05:57:45.173

Reputation: 9 469

11

In question #14739, I asked whether the product of two ideals of a commutative ring $R$ could be defined lattice-theoretically the same way the sum and intersection can. Bjorn Poonen gave a great counterexample that shows the answer is no! This supports a point fpqc had been trying to make to me earlier that the relationship between $R$ and the Zariski topology on $\text{Spec } R$ was more subtle than I had thought: in particular, it has more structure than just the Galois connection.

Qiaochu Yuan

Posted 2010-03-02T05:57:45.173

Reputation: 73 568

11

The following are, I think, the "worst possible" counterexamples in measure theory. They would benefit from a nice list of properties -- I have a feeling that I'm forgetting a lot. Feel free to improve!

The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.

Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$. ($\lambda$ is the Lebesgue measure, c denotes complement).


Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let In be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I1. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval In minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I2, and keep going forever. Every time, we add one Cantor set to A and one to B.

Now, each subinterval of [0,1] will contain one of the In-s, and therefore its intersection with both A and B has positive measure. Both A and B are countable unions of measurable sets, and therefore measurable. We are done.

Ilya Grigoriev

Posted 2010-03-02T05:57:45.173

Reputation: 2 158

1I also like the proof of the existence of the set $A$ you mentioned above using the Baire category. Consider the metric on measurable sets you get from the inclusion into L^{1}, to prove that such an $A$ exists, it suffices by Baire category to show that for any interval $I$ with rational endpoints the collection of sets $A$ which have $0<\lambda(A\cap I)<\lambda(I)$ are open and dense. But this is essentially a triviality. – Benjamin Hayes – 2011-05-02T19:53:53.767

I guess I should note that the collection of measurable sets with this metric is complete, but it is fairly clearly closed in $L^{1},$ while not as constructive as your construction it's very pain free. – Benjamin Hayes – 2011-05-02T19:55:02.317

What is a fat Cantor set? – Sune Jakobsen – 2010-04-04T15:08:31.420

@Sune: a fat Cantor set is a variation on the Cantor set that has positive measure. Just like the usual Cantor set is closed, nowhere dense, and uncountable. The difference is that the usual Cantor set has measure zero (the total length of all the intervals you remove when constructing it is 1). You construct the fat Cantor set the same way as the usual Cantor set, but you carefully vary the sizes of the intervals you remove. Apparently, it's also called the Smith-Volterra-Cantor Set. More details: http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set

– Ilya Grigoriev – 2010-04-04T20:26:16.310

2@llya The fat Cantor set is one of the great teaching examples of both analysis and topology.Most professors just go over the plain vanilla Cantor set. This is really doing the class a disservice because they don't really get the depth of the sheer diversity of pathology that can occur the real line simply by varying the details of the method of construction of the set, – The Mathemagician – 2010-07-29T18:36:07.447

11

I like the Sorgenfrey line. It's finer than the metric topology on R, and hereditarily Lindelöf, hereditarily separable, first countable, but not second countable. It's non-orderable, but generalised orderable, etc. It's a popular example for metrisation theorems, e.g. All its compact subsets are at most countable.

Henno Brandsma

Posted 2010-03-02T05:57:45.173

Reputation: 3 478

10

Volterra's function has a derivative everywhere which is bounded, discontinuous, and cannot be Riemann-integrated. It depends on the Cantor sets, of course, already mentioned.

Possible reference: Bernard R. Gelbaum, John M. H. Olmsted: Counterexamples in Analysis.

See also MO:Integrability of derivatives.

Will Jagy

Posted 2010-03-02T05:57:45.173

Reputation: 18 679

10

Two common misconceptions that students have about the concept of independence in probability are

1) Thinking "$X$ and $Y$ are independent" means "$X$ and $Y$ don't affect each other".
2) Thinking "A set of variables is independent" means the same thing as "each pair of variables are independent.

A useful counterexample that addresses both of these misconceptions at once is as follows:

Let $X$ and $Z$ be independent random variables, each equally likely to be $1$ or $-1$. Let $Y=XZ$.

In this example, changing the value of $X$ clearly changes the value of $Y$, but it does not change the distribution of $Y$. So $X$ and $Y$ are independent even though they "affect each other" in some sense. Furthermore, the variables $X, Y, Z$ are pairwise independent, but not independent.

Kevin P. Costello

Posted 2010-03-02T05:57:45.173

Reputation: 4 771

10

As a counter-example for Fatou's lemma in measure theory: strict inequality can occure! Just take the measure space $\mathbb{N}$ with the counting measure and consider the functions \begin{equation} f_n(k) = \delta_{nk} \end{equation} Then the sum of $f_n$ is always $1$ while the pointwise limit of the $f_n$ will be the zero function having zero integral. If you have this counter-example then you do not need fancy measures and integrals at al to produce examples that in Fatou's lemma strict inequality may happen...

Stefan Waldmann

Posted 2010-03-02T05:57:45.173

Reputation: 4 912

10

I occasionally use the following "counterexample" to unique factorization in Z in an introduction to math course: (1003)(1007)=(901)(1121). Once the students figure out what's going on, I think they learn something from it.

paul Monsky

Posted 2010-03-02T05:57:45.173

Reputation: 3 330

2$(17\cdot59)(19\cdot53)=(17\cdot53)(19\cdot59)$ – Akiva Weinberger – 2015-09-02T02:06:04.060

1Is the point that, despite @AkivaWeinberger's explicit factorisation, all of the multiplicands 'look prime'? (That is, is it the same (counter)example as $6\cdot35 = 14\cdot15$, but for people with some number sense?) – LSpice – 2016-05-13T04:48:42.443

10

This is an easy one, but one I've found useful in the past to keep in mind, and which I've passed on to many younger students who are new to homological algebra. These students sometimes struggle with the idea of a non-free projective module because if you're new to modules and you still think of them via analogy to vector spaces then it's natural to think direct summands of free modules should be free.

A nice counter-example to keep in mind is the ring $\mathbb{Z}/6\mathbb{Z}$ and the projective but not free module $\mathbb{Z}/3\mathbb{Z}$ (projective because $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$)

David White

Posted 2010-03-02T05:57:45.173

Reputation: 9 469

9

"Every finitely-branching tree with infinitely many nodes has an infinite branch" is constructively false, as witnessed by the following counterexample:

http://math.andrej.com/wp-content/uploads/2006/05/kleene-tree.pdf

Andrej Bauer's exposition (above) is especially nice; most textbooks take a far less direct route to the result, which makes it harder to see what's really going on past the level of "yeah, the proof is correct step-by-step."

Adam

Posted 2010-03-02T05:57:45.173

Reputation: 2 055

8

Coefficients of cyclotomic polynomials over $\mathbb{Q}$.

If you look at the factorization of $X^n-1$ over the integers, for $2 \leq n \leq 104$, you would "notice" that all nonzero coefficients of all factors are $\pm 1$. Indeed, $105$ is the first counterexample to this conjecture, with the 105th cyclotomic polynomial having coefficients of $2$ in its expansion. This can happen because $105$ has three distinct odd prime factor. The conjecture and the counterexample, however, are accessible even to high school students.

A quick Internet search suggests the following book as a reference:

McClellan, J. H. and Rader, C. Number Theory in Digital Signal Processing. Englewood Cliffs, NJ: Prentice-Hall, 1979.

I admit I have not read it - I first saw the counterexample while teaching high school, and it came up again in an advanced undergraduate course on Galois theory.

Abel Castillo

Posted 2010-03-02T05:57:45.173

Reputation: 66

8

The statement S "every injective endomap is also surjective" can be formalized in terms of second-order logic (and, of course, precisely states that the strcture in question is finite). This is a counterexample to any kind of compactness result for second-order logic, because if such a result existed, one would be able to get infinite sets satisfying S.

Akhil Mathew

Posted 2010-03-02T05:57:45.173

Reputation: 12 685

8

The Schoenflies conjecture was asserting that the two connected components of the complement of an embedded $2$-sphere $S^2$ in $S^3$ were simply connected. A kind of generalized Jordan theorem.

Antoine's necklaces gave a first counterexample, and that counterexample was reworked by Alexander to obtain the horned sphere :

enter image description here

In this counterexample, the set of singular points of the embedding is a Cantor set, so is quite big. Later, Artin and Fox developped the notion of wild arcs, and found the following simpler counterexample, where there are only two singular points :

enter image description here

few_reps

Posted 2010-03-02T05:57:45.173

Reputation: 1 360

How does Antoine's necklace give a counterexample? It's not an embedded sphere. – Akiva Weinberger – 2016-03-02T14:35:30.470

1@AkivaWeinberger Antoine's Necklace (AN) gives an example of two homeomorphic subsets of $\mathbf R^3$ whose complements are not homeomorphic (AN is a Cantor set, the complement of AN is not simply connected). I think I read in Moïse that it implies, after a detour, that the Schoenflies conjecture is false ... I'm gonna check. – few_reps – 2016-03-02T17:04:42.700

1

I found a paper that constructs "Antoine's Horned Disks", which, after identifying the boundary, creates a horned sphere. I think that that's probably what Antoine's Horned Sphere is. http://www.jstor.org/stable/2686463

– Akiva Weinberger – 2016-03-06T02:19:05.343

8

Here is a useful example of counter-examples in commutative ring theory;

Let $R=P(\mathbb{N})$ be the power set of $\mathbb{N}.$ It has a ring structure $(R, +, \times)$ where $+$ is the symmetric difference of sets and $\times$ is the intersection of sets.

Applications:

Obviously, $R$ is a commutative ring with $1$, ($\mathbb{N}$ is the $1$).

1) Let $R$ be a commutative ring with $1$ and a multiplicative closed set of $R$. If $R$ is Noetherian (Artinian) ring then $S^{-1}R$ is Noetherian (Artinian). Does the converse hold?

No, it doesn't.

Using the above example, for any prime ideal $p$ of $R$, $R_p$ (the localization at $p$) is Noetherian (Artinian) while, $R$ is not Noetherian (Artinian).

Outline:

Consider P({1}) $\subset$ P({1,2}) $\subset... $ and $P(\mathbb{N}) \supset$ P($\mathbb{N} \setminus${1}) $\supset$ P($\mathbb{N} \setminus${1,2}) $\supset ...$ showing that $R$ is neither Noetherian nor Artinian ring.

It is easy to verify that $R_p$ is isomorphic to $\mathbb{Z}/2$, hence it is both Noetherian & Artinian. (Every element of $R_p$ is either $0/1$ or a invertible.)

2) Let $R$ be an integral domain (also commutative with $1$), then for every multiplicative closed set of $R$, $S^{-1}R$ is an integral domain, hence for every $R_p.$ Does the converse hold?

By the above example, it doesn't, since $(P(\mathbb{N}),+,\times)$ is not an integral domain.

Ehsan M. Kermani

Posted 2010-03-02T05:57:45.173

Reputation: 750

2It may be worth noticing that this ring $R$ is nothing but $(\mathbb Z/2)^{\mathbb N}$ in disguise. Also, I am surprised with your statement that localizations $R_p$ are all isomorphic to $\mathbb Z/2$. – ACL – 2011-03-17T09:10:06.733

@ACL: Good point. Is it possible to understand $Spec(R)$ in this example?

The point is that $P(A)$ where $A \subset \mathbb{N}$ is an ideal but not all ideals are being as such! – Ehsan M. Kermani – 2011-03-17T10:16:51.673

4The prime ideals in this ring are the complements of the ultrafilters on $\mathbb N$, so the spectrum is the Stone-Cech compactification of the discrete space $\mathbb N$. – Andreas Blass – 2011-03-17T13:53:12.763

7

Any classical counter-example to inversion of a limit and an integral, $f_n:[0,1[\to\mathbb{R} ; x\mapsto n^2 x^n$ say. Basic, but important to motivate the dominated convergence theorem.

Benoît Kloeckner

Posted 2010-03-02T05:57:45.173

Reputation: 10 345

6

Assume given three projective systems $\{A_n,\alpha_{nm}\}_{n\in\mathbb{N}}$, $\{B_n,\beta_{nm}\}_{n\in\mathbb{N}}$ and $\{C_n,\kappa_{nm}\}_{n\in\mathbb{N}}$ of abelian groups (modules over some ring would equally do), endowed with arrrows $$ 0\rightarrow A_n\xrightarrow{f_n}B_n\xrightarrow{g_n}C_n\rightarrow 0 $$ making the above sequences exact for every $n$ and satisfying the commutativity conditions $\beta_{nm}\circ f_n=f_m\circ\alpha_{nm}$ and $\kappa_{nm}\circ f_n=f_m\circ\beta_{nm}$. Then one can form the projective limits of the system to find a sequence $$ 0\rightarrow \varprojlim A_n\xrightarrow{f}\varprojlim B_n \xrightarrow{g}\varprojlim C_n $$ and a classical result says that, in order for this sequence to be right-exact, one needs the system $A_n$ to be stationary - meaning that $\alpha_{nm}(A_n)=\alpha_{n'm}(A_{n'})\subseteq A_m$ for all $n,n'\gg m$.

A classical counterexample showing the necessity of this condition is to take $A_n=p^n\mathbb{Z}$ with $\alpha_{nm}$ given by inclusions, $B_n=\mathbb{Z}$ for all $n$ with identity maps $\beta_{nm}=\mathrm{id}$, and $C_n=\mathbb{Z}/p^n\mathbb{Z}$ with the obvious maps. The system $A_n$ is non-stationary because the image of $A_n$ in $A_m$ is $p^n\mathbb{Z}\subseteq p^m\mathbb{Z}$ which becomes smaller and smaller as $n\rightarrow \infty$: the corresponding sequence of projective limits is $$ 0\rightarrow 0\rightarrow \mathbb{Z}\rightarrow\mathbb{Z}_p $$ which is clearly not right exact.

[Later remark]: After typing all down, I remarked that everything can be found in Wikipedia at http://en.wikipedia.org/wiki/Inverse_limit Moreover, the stationary condition quoted above, usually referred to as Mittag-Leffler condition, is enough to prove right-exactness of $\varprojlim$ in Ab, but there is a counterexample due to Deligne and Neeman showing that in other categories this is not enough, see http://www.springerlink.com/content/aeem2yx884nnufxn/

Filippo Alberto Edoardo

Posted 2010-03-02T05:57:45.173

Reputation: 3 518

6

Ackermann function (http://en.wikipedia.org/wiki/Ackermann_function) defined as

$ A(m, n) = \begin{cases} n+1 & \mbox{if } m = 0 \\ A(m-1, 1) & \mbox{if } m > 0 \mbox{ and } n = 0 \\ A(m-1, A(m, n-1)) & \mbox{if } m > 0 \mbox{ and } n > 0. \end{cases} $

is total recursive but not primitive recursive. To see this we could prove by induction on the complexity of primitive recursive functions that each primitive recursive function is eventually dominated by this function (we need a bit coding to keep the number of arguments consistent). Essentially this function manages to capture the ``fast-growing'' property. Note that the index set for recursive functions is not recursive while that for primitive recursive function is.

Jing Zhang

Posted 2010-03-02T05:57:45.173

Reputation: 1 513

6

The elliptic curve 960d1 in Cremona's tables is the smallest conductor example of an optimal elliptic curve with nontrivial Shafarevich-Tate group which is isogenous to an elliptic curve with trivial Shafarevich-Tate group.

Jamie Weigandt

Posted 2010-03-02T05:57:45.173

Reputation: 2 373

1I had been concerned about wether this was proven to be the smallest example. It is now, thanks to the work of Robert Miller. – Jamie Weigandt – 2011-04-30T02:38:15.093

6

Here is some simple counterexample in commutative algebra, which I found really cute when I first meet it:

Let $k$ be a field, $A = k[X_{1},X_{2},X_{3}\ldots],$ $I = (X_{1}, X_{2}^{2}, X_{3}^{3},\ldots)$ and $R = A/I.$ Then $\text{Spec}(R)$ consists of one point (because $\text{rad}(I)$ is maximal ideal of $A$); in particular $\text{Spec}(R)$ is a noetherian space, and $\dim R = 0$; although $R$ is not noetherian ring (since $\text{nil}(R)^{n}\neq 0$ for every $n$).

ifk

Posted 2010-03-02T05:57:45.173

Reputation: 504

I don't know; I think it's pretty intuitive that there exist local rings that aren't Noetherian. Nothing in the definition of a local ring suggests that they need be Noetherian. – Qiaochu Yuan – 2010-04-04T21:21:22.323

2Nevertheless, I think that it's not obvious that there exist commutative rings with only one prime (not only with one maximal) ideal that are not noetherian. – ifk – 2010-04-04T21:35:51.823

3@ifk: There are simpler examples of that: Consider the direct sum $R=k\oplus V$ of a field $k$ and an infinite dimensional vector space $V$, made into a ring so that $V$ is an ideal which squares to zero, $k$ and $V$ multiply as you expect, and $k$ is a subring (this is called a trivial extension, in some contexts) Then $R$ is commutative, has only one prime, and it is not noetherian. – Mariano Suárez-Álvarez – 2010-04-05T06:05:04.390

Nice, thank You Mariano. However I don't think it's really much simpler than above. – ifk – 2010-04-05T09:47:48.070

6

The matrix pencil $$ \left\{ \begin{bmatrix} 0 & 1 & x\\ 1 & 0 & 0\\ x & 0 & 0 \end{bmatrix} : x \in \mathbb{R} \right\}. $$ The matrices composing it are all singular, but they have no common left or right kernel (which is a property that one expects when first diving into the theory of matrix pencils). Singular pencils are difficult (or impossible) to handle for algorithms to solve generalized eigenvalue problems. For instance, Matlab's eig([0 1 0; 1 0 0; 0 0 0],[0 0 1; 0 0 0; 1 0 0]) returns 0 NaN 0 instead of something like NaN NaN NaN which would make more sense (no zero eigenvalues here), since the algorithm is not designed to handle this kind of singular problems.

Federico Poloni

Posted 2010-03-02T05:57:45.173

Reputation: 11 194

6

Rotations $\rho_\alpha$ of the unit circle by an angle $2\pi\alpha$ are nice examples in the theory of discrete dynamical systems.

If $\alpha=m/n$ is rational, then every point on the circle is periodic of prime period $n$ for $\rho_\alpha$, but has no fixed points. This shows that Sharkowskii's theorem does not hold in general for functions continuous $f\colon X\to X$ if $X$ is not the real line or an interval of the real line.

If $\alpha$ is irrational, then the orbit under $\rho_\alpha$ of every point of the circle is dense, but $\rho_\alpha$ has nor sensitive dependence on initial conditions, and in particular is not caotic.

Julián Aguirre

Posted 2010-03-02T05:57:45.173

Reputation: 1 448

6

The 5-cycle $C_5$ is a great counterexample. It's the smallest imperfect graph, it's self-complementary, it has chromatic number $>\Delta$, it has no stable set meeting every maximum clique and yet satisfies $\omega = \frac{2}{3}(\Delta+1)$, it has chromatic number $> \frac 1 2 (\Delta+\omega+1)$, meaning that Reed's $\chi, \omega, \Delta$ conjecture is somehow tight.

And when you blow up each vertex into a clique or stable set of size $k$, the fun continues. For $k=3$ this gives you Catlin's counterexample to Hajos' Conjecture.

Andrew D. King

Posted 2010-03-02T05:57:45.173

Reputation: 1 321

6

I'm shocked that noone has mentioned the Quaternion group! This thing is a counterexample to lots of basic questions you'd come up with while learning (finite) group theory.

For example (although not really a counterexample to a specific question), if you know the semidirect product construction and Sylow theorems and are trying to classify groups of low order, the quaternion group is the first group you can't construct as a semidirect product of cyclic groups. This can be an entry point for the extension problem for groups and cohomology of groups.

Jon Bannon

Posted 2010-03-02T05:57:45.173

Reputation: 3 591

6

$\textbf{Algebra.}$

  • The symmetric group $S_{3}$ is the first $\text{non-abelian}$ group and also this group has a fascinating property that $S_{3} \cong \mathscr{I}(S_{3})$ where $\mathscr{I}$ denotes the $\text{Inner - Automorphism}$ group.

  • Example of a group which is $\textbf{isomorphic}$ to it's proper subgroup. $\mathsf{Answer:}$ Take $G=(\mathbb{Z},+)$ and take $H= 2\mathbb{Z}$. Then $G \cong H$.

  • Example of a free module in which a linearly independent subset cannot be extended to a basis. $\textbf{Answer.}$ As a $\mathbb{Z}$ module $\mathbb{Z}$ is free with basis $\{1\}$ and $\{-1\}$. Now $\{2\}$ is linearly independent over $\mathbb{Z}$. Note that $2$ cannot generate $\mathbb{Z}$ over $\mathbb{Z}$. If at all there is a basis $\mathscr{B}$ containing $2$, $\mathscr{B}$ should have atleast one more element, say $b$. We then have $b\cdot 2 - 2\cdot b =0$, i.e $\{2,b\}$ is linearly dependent subset of $\mathscr{B}$ which is absurd.

$\textbf{Analysis.}$

  • The function defined by $f(x) = x^{2} \cdot \sin\frac{1}{x}$ for $x \neq 0$ and $f(x) =0$ for $x=0$. This is example of a function whose derivatives are not continuous.

  • Set that is not Lebesgue measurable. Example given by Vitali.

crskhr

Posted 2010-03-02T05:57:45.173

Reputation: 1 992

5

My favorite counter-example is given in the short paper, "Almost Commuting Unitaries," by R. Exel and T. Loring.

Here is a little background. Two $n \times n$ matrices $A$ and $B$ are said to be "almost-commuting" if there commutator, $[A, B]$, is small in some matrix norm. In the paper, the authors exhibit a family of unitary matrices, $U_n$ and $V_n$ that "almost-commute" in the sense that given $\epsilon > 0$ there exists an $N \in \mathbb{N}$ with $|| [U_n, V_n] || < \epsilon$ for all $n \geq N$, yet for any commuting $n \times n$ matrices, $X, Y$ $(XY = YX)$ there exists an absolute constant $C > 0$ such that $\max(||X - U_n||, ||Y - V_n||) > C > 0$. This was one of the first counter-examples in a research paper that I understood because the authors method of proof is very elementary. The most technical fact used is that the winding number of a closed curve around the origin is a homotopy invariant.

Mustafa Said

Posted 2010-03-02T05:57:45.173

Reputation: 973

4

A nice counterexample to the statement "$L^p$ convergence to $0$ implies pointwise a.e. convergence to $0$" is obtained by taking characteristic functions of length $\frac{1}{n}$ wrapping around the interval $[0,1]$. These integrate to $\frac{1}{n}$, but converge nowhere to $0$ because the harmonic series diverges.

A counterexample to the converse is easier: just take $f_n = n(n+1)\chi_{[\frac{1}{n+1},\frac{1}{n}]}$. These integrate to $1$ and converge everywhere to $0$.

Connor Mooney

Posted 2010-03-02T05:57:45.173

Reputation: 1 921

I'm sorry, could I ask you to clarify what you mean by "characteristic functions of length $1/n$ wrapping around the interval $[0, 1]$". I ask as I thought $L^p$ convergence to zero did imply pwae convergence to $0$, so I would like to understand your example. – Arch Stanton – 2012-05-19T13:53:51.723

1$\chi{[0,1/2]}, \chi{[1/2,5/6]}, \chi{[5/6,1] \cup [0,1/12]}, \chi{[1/12,17/60]}...$. – Douglas Zare – 2012-07-10T04:58:08.393

2I should also mention here that $L^p$ convergence of ${f_k}$ to 0 implies that a subsequence converges pointwise a.e. to zero. To see this take a subsequence with $\int |f_k|^p < 2^{-k}$ (or any summable series) and use the monotone convergence theorem to conclude that $\int \sum |f_k|^p < \infty$. – Connor Mooney – 2012-08-02T12:10:43.067

3

In topology, The comb space is an example of a path connected space which is not locally path connected. see http://en.wikipedia.org/wiki/Comb_space.

GA316

Posted 2010-03-02T05:57:45.173

Reputation: 179

3

I am quite keen of a counterexample in Strichartz estimates by Thomas Wolff that I think satisfies the requirements in your question. The problem was whether or not there are $L^p$ Strichartz estimates that lose no derivatives. Wolff gave a counterexample that, assuming the best control possible, then this is not possible for $p>2$. It uses the result by Charles Fefferman that the disk multiplier $S_1$ is not bounded on $L^p,~p\neq 2$. This is brilliantly explained by Professor Tao in the Notice of the AMS article "From Rotating Needles to Stability of Waves: Emerging Connections Between Combinatorics, Analysis, and PDE". It goes (more-or-less) like this:

Let $B(0,1)$ denote the unit ball of centre $0$ and radius $1$. Wolff showed that the inequality \begin{equation*} ||u||_{L^p([1,2]\times \mathbb{R}^n)}\leq C||f||_{L^{\infty}(B(0,1))} \end{equation*} fails, for $f$ bounded on $B(0,1)$.

Wolff used a mixture of Kakeya tube constructions, and the manipulation of specially constructed waves consisting of bump and exponential functions adapted to the tubes.

George Simpson

Posted 2010-03-02T05:57:45.173

Reputation: 322

3

From an earlier post: "The 8-element quaternion group. It can't be reconstructed from its character table (D_4 has the same one), and every subgroup is normal but it's not abelian."

Although the character tables for the dihedral group D of order 8 and the quaternion group Q of order 8 may seem the same, they are not. Using Adams operations on the representation rings for D and Q, it is possible to show that these representation rings are different as rings with operations (either lambda or Adams operations). These Adams operations are defined in a paper by Aityah and Tall, where it is shown how to calculate them directly from character tables.

Paul Pearson

Posted 2010-03-02T05:57:45.173

Reputation: 151

7It can't be possible to define lambda operations directly from the character table; you need to know some of the multiplication table as well. – Qiaochu Yuan – 2011-07-31T18:07:14.690

2

Another one of my favorite counter examples is $2\mathbb{Z}$ which is a RNG, or a ring without identity.

Mustafa Said

Posted 2010-03-02T05:57:45.173

Reputation: 973

2

My favorite example in discrete mathematics is the sequence $1,2,4,8,16,31,..$. That is, number of regions in a circle after drawing $n$ chords.

It shows that a simple pattern might be wrong, and that we do need formal proofs, no matter how many examples we've checked.

Per Alexandersson

Posted 2010-03-02T05:57:45.173

Reputation: 6 151

I feel like this is a also good example of the power of intuition - we 'should' expect this sort of configuration to be polynomially sized, because of the geometry, and so one should innately be suspicious that the natural pattern can continue indefinitely. – Steven Stadnicki – 2018-02-05T20:17:46.587

1

I like "the deleted Tychonov plank" which is described in "Counterexamples in Topology".

This space provide us a pure algebraic counterexample:

A commutative ring is called a $Z$-ring if all its elements are zero divisors. At first glance, it seems that an extension of a $Z$-ring by a $Z$-ring is necessarily a $Z$ ring. But this space provides a counterexample. The reason is explained here.

http://arxiv.org/abs/1307.5836

To be honest, I spent a few weeks to give a (positive) proof for this ring-extension statment, but finaly I found this counterexample in the book "Counterexamples in Topology" which excited me a lot.

As a consequence of this post we ask:

Are there two $Z$-rings $R_{1}$ and $R_{2}$ such that for every short ring exact sequence $$0\to R_{1} \to S \to R_{2} \to 0$$ $S$ must be a $Z$-ring?

Ali Taghavi

Posted 2010-03-02T05:57:45.173

Reputation: 234

1

I am surprised no one mentioned the unilateral shift on $l^2 ({\bf N})$, that is, $Te_i = e_{i+1}$ where $\{e_i\}$ is the standard orthonormal basis.

It provides examples for the following.

(a) a ring with elements $x,y$ such that $xy = 1 \neq yx$;

(b) an isometry from a Hilbert space to itself that is not a unitary;

(c) a one to one BLT that is not invertible;

(d) its adjoint is an onto BLT with nontrivial kernel;

and lots of other stuff (for example, its spectrum).

David Handelman

Posted 2010-03-02T05:57:45.173

Reputation: 3 247

Where is the ring? – Gerry Myerson – 2017-02-09T22:01:34.230

2@GerryMyerson For (a) one doesn't have to be so fancy, but it could be the ring of bounded linear maps taking $l^2$ to itself, with multiplication given by composition. Here $y$ would be given by $T$, and $x$ by $e_0 \mapsto e0$ and $e{i+1} \mapsto e_i$. – Todd Trimble – 2017-02-09T22:46:03.670

Easier is $T^ : ei \mapsto e{i-1}$ if $i > 1$, and $T^e_1 = 0$ (there is no $e_0$). – David Handelman – 2017-02-10T14:08:59.953

Well, I say there is an $e_0$ because $0$ is a natural number (belongs to $\mathbf{N}$). But it's not important here. – Todd Trimble – 2017-02-11T14:13:54.393

This boils down to the use of ${\bf N}$ to denote the positive integers (my use) or the nonnegative integers (your use); I prefer using $\bf Z^+$ for the latter, which avoids some ambiguity ... (and although it is nonstandard, $\bf Z^{++}$ for the positive integers avoids more ambiguity). – David Handelman – 2017-02-11T14:21:54.653

1

One might expect that for any function $f$ in the Hilbert space $L^2$ with orthonormal basis $\{ f_i \}$, the generalized Taylor series $$\sum_{j=1}^\infty \langle f, f_j \rangle f_j$$ converges pointwise to $f$ almost everywhere. Carleson's theorem gives that this is true for the standard Fourier basis of $L^2([0,1])$, but there exist uncountably many pairs of functions $g$ and orthonormal (ordered) bases $\{ f_i \}$ such that the generalized Fourier series above diverges pointwise almost everywhere, as discussed here.

tparker

Posted 2010-03-02T05:57:45.173

Reputation: 121

-10

the example which shows that exp(zw) is not equal to exp (exp(z)^ w) the carrot sign means raised to the power another one a continuous function of a complex variable need not have primitive in a region.the example is f(z) = square ( | z| ).

Anil P

Posted 2010-03-02T05:57:45.173

Reputation: 56

why -2c marks these are simple but important counter examples – Anil P – 2011-07-01T16:53:08.163

What does exp(exp(z), w) mean? – LSpice – 2016-05-13T04:56:30.747