## Why worry about the axiom of choice?

157

148

As I understand it, it has been proven that the axiom of choice is independent of the other axioms of set theory. Yet I still see people fuss about whether or not theorem X depends on it, and I don't see the point. Yes, one can prove some pretty disturbing things, but I just don't feel like losing any sleep over it if none of these disturbing things are in conflict with the rest of mathematics. The discussion seems even more moot in light of the fact that virtually none of the weird phenomena can occur in the presence of even mild regularity assumptions, such as "measurable" or "finitely generated".

So let me turn to two specific questions:

If I am working on a problem which is not directly related to logic or set theory, can important mathematical insight be gained by understanding its dependence on the axiom of choice?

If I am working on a problem and I find a two page proof which uses the fact that every commutative ring has a maximal ideal but I can envision a ten page proof which circumvents the axiom of choice, is there any sense in which my two page proof is "worse" or less useful?

The only answer to these questions that I can think of is that an object whose existence genuinely depends on the axiom of choice do not admit an explicit construction, and this might be worth knowing. But even this is largely unsatisfying, because often these results take the form "for every topological space there exists X..." and an X associated to a specific topological space is generally no more pathological than the topological space you started with.

36For bonus points, provide an answer to the question comprehensible by an undergraduate! – Mariano Suárez-Álvarez – 2010-04-29T03:01:04.173

3I think even non-logicians should make the effort to properly understand the second part of Hamkins answer. – None – 2010-04-29T04:21:30.080

2I like the question, but not the first paragraph, which is a bit too discussion-y/argumentative for my taste. – Theo Johnson-Freyd – 2010-04-29T04:36:31.097

1I think we should start caring about the axiom of regularity instead. That's the one that is not necessary for quite a large part of modern mathematics. – Zsbán Ambrus – 2010-04-29T09:04:47.237

9Zsbán, but isn't it also true that the relative consistency of Regularity is very easy to prove, in a way that seems to assuage lingering set-theoretic worries about it? Namely, every set-theoretic universe without Regularity contains a large subuniverse with Regularity, the class of well-founded sets. So unless you prefer ill-foundedness for some reason (and some do), it seems unproblematic to assume this axiom. The situation with the Replacement axiom, in contrast, is very different. – Joel David Hamkins – 2010-04-29T14:27:07.010

5"What me worry?" – Simon Thomas – 2010-05-03T00:10:39.190

Joel, isn't that just as true about choice? Every set-theoretic universe without Choice contains a subuniverse with Choice, namely the constructible universe. – Mike Shulman – 2010-05-03T13:59:33.863

2Mike: I don't think it's true with quite as strong a sense of "sub-universe". If we start without regularity, the inclusion of the class WF of well-founded sets into the full universe V should be logical in most senses (I'm not quite sure which, but it shouldn't be hard to make some precise) so WF is closed under all the constructions we use. On the other hand, the inclusion of L into V is not logical: power sets, function spaces, infinite limits etc. are computed differently in L, and the truth of a lot of bounded statements can change. – Peter LeFanu Lumsdaine – 2010-05-03T17:57:26.820

1Mike, I agree, and this is part of why we should feel safe assuming AC. (But I wouldn't describe L as "large", since one loses too many large cardinals going down to L.) The situation with Replacement, however, is totally different, since the strength of ZF or ZFC is far stronger than ZF-Replacement. – Joel David Hamkins – 2010-05-11T20:19:07.360

110

The best answer I've ever heard --- and I think I heard it here on MathOverflow from Mike Shulman, which suggests that this question is roughly duplicated somewhere else --- is that you should care about constructions "internal" to other categories:

1. For many, many applications, one wants "topological" objects: topological vector spaces, topological rings, topological groups, etc. In general, for any algebraic gadget, there's a corresponding topological gadget, by writing the original definition (a la Bourbaki) entirely in terms of sets and functions, and then replacing every set by a topological space and requiring that every function be continuous.
2. A closely related example is that you might want "Lie" objects: sets are replaced by smooth manifolds and functions by smooth maps.
3. Another closely related example is to work entirely within the "algebraic" category.

In all of these cases, the "axiom of choice" fails. In fact, from the internal-category perspective, the axiom of choice is the following simple statement: every surjection ("epimorphism") splits, i.e. if $f: X\to Y$ is a surjection, then there exists $g: Y \to X$ so that $f\circ g = {\rm id}_Y$. But this is simply false in the topological, Lie, and algebraic categories.

This leads to all sorts of extra rich structure if you do algebra internal to these categories. You have to start thinking about bundles rather than products, there can be "anomalies", etc.

Update:

In the comments, there was a request for a totally explicit example, where Axiom of Choice is commonly used but not necessary. Here's one that I needed recently. Let $\mathcal C$ be an abelian tensor category, by which I mean that it is abelian, has a monoidal structure $\otimes$ that is biadditive on hom-sets, and that has a distinguished natural isomorphism $\text{flip}: X\otimes Y \overset\sim\to Y\otimes X$ which is a "symmetry" in the sense that $\text{flip}^2 = \text{id}$. Then in $\mathcal C$ is makes sense to talk about "Lie algebra objects" and "associative algebra objects", and given an associative algebra $A$ you can define a Lie algebra by "$[x,y] = xy - yx$", where this is short-hand for $[,] = (\cdot) - (\cdot \circ \text{flip})$ — $x,y$ should not be read as elements, but as some sort of generalization. So we can makes sense of the categories of $\text{LIE}_{\mathcal C} =$"Lie algebras in $\mathcal C$" and $\text{ASSOC}_{\mathcal C} =$"associative algebras in $\mathcal C$", and we have a forgetful functor $\text{Forget}: \text{ASSOC}_{\mathcal C} \to \text{LIE}_{\mathcal C}$.

Then one can ask whether $\text{Forget}$ has a left adjoint $U: \text{LIE}_{\mathcal C} \to \text{ASSOC}_{\mathcal C}$. If $\mathcal C$ admits arbitrary countable direct sums, then the answer is yes: the tensor algebra is thence well-defined, and so just form the quotient as you normally would do, being careful to write everything in terms of objects and morphisms rather than elements. In particular, if $\mathfrak g \in \text{LIE}_{\mathcal C}$, then $U\mathfrak g \in \text{ASSOC}_{\mathcal C}$ and it is universal with respect to the property that there is a Lie algebra homomorphism $\mathfrak g \to U\mathfrak g$.

Let's say that $\mathfrak g$ is representable if the map $\mathfrak g \to U\mathfrak g$ is a monomorphism in $\text{LIE}_{\mathcal C}$. By universality, if there is any associative algebra $A$ and a monomorphism $\mathfrak g \to A$, then $\mathfrak g \to U\mathfrak g$ is mono, so this really is the condition that $\mathfrak g$ has some faithful representation. The statement that "Every Lie algebra is representable" is normally known as the Poincare-Birkoff-Witt theorem.

The important point is that the usual proof — the one that Birkoff and Witt gave — requires the Axiom of Choice, because it requires picking a vector-space basis, and so it works only when $\mathcal C$ is the category of $\mathbb K$ vector spaces for $\mathbb K$ a field, or more generally when $\mathcal C$ is the category of $R$-modules for $R$ a commutative ring and $\mathfrak g$ is a free $R$-module, or actually the proof can be made to work for arbitrary Dedekind domains $R$. But in many abelian categories of interest this approach is untenable: not every abelian category is semisimple, and even those that are you often don't have access to bases. So you need other proofs. Provided that $\mathcal C$ is "over $\mathbb Q$" (hom sets are $\mathbb Q$-vector spaces, etc.), a proof that works constructively with no other restrictions on $\mathcal C$ is available in

• Deligne, Pierre; Morgan, John W. Notes on supersymmetry (following Joseph Bernstein). Quantum fields and strings: a course for mathematicians, Vol. 1, 2 (Princeton, NJ, 1996/1997), 41--97, Amer. Math. Soc., Providence, RI, 1999. MR1701597.

They give a reference to

• Corwin, L.; Ne'eman, Y.; Sternberg, S. Graded Lie algebras in mathematics and physics (Bose-Fermi symmetry). Rev. Modern Phys. 47 (1975), 573--603. MR0438925.

in which the proof is given when $\mathcal C$ is the category of modules of a (super)commutative ring $R$, with $\otimes = \otimes_R$, and, importantly, $2$ and $3$ are both invertible in $R$. [Edit: I left a comment July 28, 2011, below, but should have included explicitly, that Corwin--Ne'eman--Sternberg require more conditions on $\mathcal C$ than just that $2$ and $3$ are invertible. Certainly as stated "PBW holds when $6$ is invertible" is inconsistent with the examples of Cohn below.]

Finally, with $R$ an arbitrary commutative ring and $\mathcal C$ the category of $R$-modules, if $\mathfrak g$ is torsion-free as a $\mathbb Z$-module then it is representable. This is proved in:

• Cohn, P. M. A remark on the Birkhoff-Witt theorem. J. London Math. Soc. 38 1963 197--203. MR0148717

So it seems that almost all Lie algebras are representable. But notably Cohn gives examples in characteristic $p$ for which PBW fails. His example is as follows. Let $\mathbb K$ be some field of characteristic $p\neq 0$; then in the free associative algebra $\mathbb K\langle x,y\rangle$ on two generators we have $(x+y)^p - x^p - y^p = \Lambda_p(x,y)$ is some non-zero Lie series. Let $R = \mathbb K[\alpha,\beta,\gamma] / (\alpha^p,\beta^p,\gamma^p)$ be a commutative ring, and define $\mathfrak g$ the Lie algebra over $R$ to be generated by $x,y,z$ with the only defining relation being that $\alpha x = \beta y + \gamma z$. Then $\mathfrak g$ is not representable in the category of $R$-modules: $\Lambda_p(\beta y,\gamma z)\neq 0$ in $\mathfrak g$, but $\Lambda_p(\beta y,\gamma z)= 0$ in $U\mathfrak g$.

Could you sketch what you mean by anomalies' ? Is the idea simply that we have more interesting behaviour in objects internal to these categories than in the similarly defined objects internal tosimpler' categories, or is there some more precise notion being hit upon (in particular, I would love to know if you're making reference to the anomalies in quantum theory !) ? – Arnav Tripathy – 2011-07-28T15:21:29.097

I think I misread something in the D+M paper's description of the CNS paper: they do seem to put in more conditions than I remembered. – Theo Johnson-Freyd – 2011-07-28T17:51:25.130

@Arnav: I was not trying to say anything completely precise about quantum theory, but definitely I get the word "anomaly" from some version of physics. The idea is that often physicists want to choose coordinates, and then any space of sections of a bundle is (locally) a space of maps from the base to a fixed fiber. But globally in many bundles this fails. An example is that on spaces with interesting topology there can be magnetic fields (satisfying Maxwell equations) for which there is not a globally-defined "magnetic potential". It exists as a section of a $U(1)$-bundle with (continued) – Theo Johnson-Freyd – 2011-07-28T17:55:51.630

(continuation) connection, but not the trivial bundle. This is not a problem classically, where only the E+M fields are physical, but quantum-mechanically a little bit of the magnetic potential is also physical (c.f. Aharonov–Bohm effect). So this is one example, and probably the example I was thinking of. (My impression is that the physicists' word "anomaly" just means "using local coordinates that do not extend to global coordinates" or "pretending something is a principal bundle when it is not (canonically) so".) – Theo Johnson-Freyd – 2011-07-28T18:01:19.670

The axiom of choice is equivalent to the Hahn-Banach theorem + the Krein-Milman theorem. These are theorems about topological vector spaces. Does this go against this answer which suggests avoiding choice in other to internalize set-theoretic arguments to the topological category? – None – 2013-08-07T06:12:42.760

@ColinTan: Certainly there are theorems about topological things or about abelian groups that are equivalent to Choice. But those arguments are necessarily made outside the category of interest. – Theo Johnson-Freyd – 2013-08-07T13:02:01.537

19I think this is the real reason to avoid AC when you can: even if it's true for boring old sets, in lots of other natural situations it's just false. This is the same point that Harry was making, but I think it's seeing the examples that really drives the point home. – Mike Shulman – 2010-04-29T05:14:33.753

5Do you have a "shocking" example to drive the point home? Ideally, something that you would usually use AC to do in Set, but can be done in some non-AC topos with useful consequences. – François G. Dorais – 2010-04-29T05:42:14.287

6From what I understand the axiom of choice implies the law of the excluded middle. Synthetic differential geometry depends on the "no mans land" around 0 functioning as an infinitesimal interval - a tiny interval where all of the points are not not zero. So SDG couldn't get of the ground if you insist on AC. – Steven Gubkin – 2010-04-29T14:05:50.133

@François: well, the immediate examples for me are the opposite direction from what you're looking for, e.g. that not every exact sequence of modules splits. Something for which it is convenient but not necessary to use choice? Often there are statements about bundles that are reasonably trivial for products (trivial bundles) but are still true, if nontrivial, for bundles. – Theo Johnson-Freyd – 2010-04-29T16:43:01.623

I normally don't buy category theory. But this is a notable exception. – John Jiang – 2015-01-02T17:21:03.737

6This is an interesting and persuasive answer. Worrying about the axiom of choice in some sense corresponds to worrying about whether or not a construction is canonical, which can certainly be a useful thing to worry about.

I liked lots of the other answers too, but this one is most convincing for me personally. – Paul Siegel – 2010-05-01T10:52:51.653

1This looks like a very nice example Theo! Thanks! – François G. Dorais – 2010-05-01T17:35:56.257

I guess another example would be writing out what essentially surjective and fully faithful mean for a topological category. This is different from having an equivalence- this leads to the idea of Morita equivalence in the case of topological groupoids. – David Carchedi – 2010-05-13T21:48:00.923

2

@JohnJiang You don't normally buy category theory? Good! http://facultypages.ecc.edu/alsani/ct99-00%288-12%29/msg00279.html

– Todd Trimble – 2016-03-31T01:36:38.990

In Bourbaki definition of manifold the following order of notions is used: set, manifold, topological space. This is contrary to the definition of manifold in some other textbooks, where another order of notions is used: set, topological space, manifold. – Vladimir – 2017-01-12T22:35:15.997

1In case anyone still reads this old post, I should have emphasized something that bothers me in my write-up. Namely, I've read both the Cohn paper and the Deligne-and-Morgan paper, but not the referenced paper by Corwin, Ne'eman, and Sternberg. But the CNS paper cannot be consistent with Cohn's paper, and I believe Cohn: Cohn gives a counterexample in algebras over a field of characteristic $p \neq 0$, and $2,3$ are invertible if $p\geq 5$. – Theo Johnson-Freyd – 2010-10-12T05:07:09.210

115

How I Learned to Stop Worrying and Love the Axiom of Choice

The universe can be very a strange place without choice. One consequence of the Axiom of Choice is that when you partition a set into disjoint nonempty parts, then the number of parts does not exceed the number of elements of the set being partitioned. This can fail without the Axiom of Choice. In fact, if all sets of reals are Lebesgue measurable, then it is possible to partition $2^{\omega}$ into more than $2^{\omega}$ many pairwise disjoint nonempty sets!

4Did you set up the account just to post this? – Niemi – 2013-10-10T16:19:36.390

8@Niemi: He changed it when he became a logician. Used to be Merkwürdigauswahl. – Torsten Schoeneberg – 2014-01-30T23:43:58.307

2Do you have a good citation for that? – Dan Piponi – 2010-04-29T04:57:04.940

28If every set of reals is Lebesgue measurable then $\omega_1 \nleq 2^{\omega}$, but then you can partition $2^{\omega}$, or rather $\mathcal{P}(\omega\times\omega)$, into $\omega_1+2^{\omega} > 2^{\omega}$ pieces by putting two wellorderings of $\omega$ in the same piece iff they have the same order type, and all non-wellorderings into singleton pieces. – Dr Strangechoice – 2010-04-29T05:07:00.053

1Very nice ! – Joel David Hamkins – 2010-04-29T05:14:36.183

8"If every set of reals is Lebesgue measurable then $\omega_1 \not\le 2^\omega$." I feel like I'm missing something that's supposed to be clear ... could somebody explain why this line is true? – Anton Geraschenko – 2010-04-29T05:41:01.073

19@Anton: This is not at all obvious, it's a theorem of Jean Raisonnier from the 1980's. – François G. Dorais – 2010-04-29T14:34:38.877

@François: Thanks! – Anton Geraschenko – 2010-04-30T03:48:14.043

17The statement that either there is a nonmeasurable set or R has a surjection to a larger set, is Sierpinski: Sur une proposition qui ..., Fundamenta Math, 34(1947), 157-162. It is also in my book (with Vilmos Totik) Problems and Theorems in Classical Set Theory, Springer, 2006, page 127. – Péter Komjáth – 2010-07-02T15:57:29.523

86

Yes, many people continue to fuss about the Axiom of Choice.

At least part of the explanation for why people continue to fuss as they do over the Axiom of Choice is surely the historical fact that there was a period of several decades during which the axiom was not known to be relatively consistent with the other axioms of set theory. It was after all not until 1938 that Goedel proved the relatively consistency of ZFC over ZF, using the constructible universe, and several more decades passed until Paul Cohen completed the independence proof by proving that ¬AC is also relatively consistent with ZF, using the method of forcing in 1962. It was during these intermediate times, and especially the time before 1938 when the axiom was not known to be consistent, that the increasingly bizarre consequences of AC were being discovered, and so the habit naturally developed to pay close attention to when the axiom was used. This habit surely lessened after the independence results, but it was not dropped by everyone. And so today mathematics is populated by large numbers of mathematicians like yourself (and perhaps myself), who freely use AC without worry, and who may even find the possibilities occuring in non-AC situations, such as infinite Dedekind finite sets, to be even weirder than the supposed non-regularities of AC, such as the existence of non-measurable sets.

Yet, even though I largely agree with the feeling you indicate in your question, there is still some reason to pay attention to AC. First, in mathematical situations where one can prove the existence of a mathematical structure without AC, then important consequences often follow concerning the complexity of that structure. An explicit construction, even if more complicated that a pure existence proof from AC, often carries with it computational information concerning the nature of the object constructed, such as whethere it is analytic or Borel or $\Delta^1_2$, and so on, and these complexity issues can affect other arguments concerning measurability and whatnot. That is, by nature the non-AC constructions are more explicit and these more explicit argument often carry more information.

But second, there remain certain parts of set-theoretic investigation that only make sense in non-AC contexts. The Axiom of Determinacy, for example, stands in contradiction with the Axiom of Choice but nevertheless contains some fascinating, profound mathematical work, pointing to a kind of mathematical paradise, in which every set of reals is Lebesgue measurable, every set has the property of Baire and the perfect set property. This axiom leads to an alternative vision of what set theory could be like. The possibilities of AD place limitations on what we can expect to prove in ZFC, in part because we expect that there are set-theoretic universe close to our own where AD holds. That is, we are interested in AD even if we believe fundamentally in AC, because we can construct the universe L(R), where AD could hold, even if AC holds in the outer universe V. In order to understand L(R), we need to know which theorems we can rely on there, and so we need to know where we used AC.

4Your first reason is the one I think one can sell. – Mariano Suárez-Álvarez – 2010-04-29T03:56:29.180

Yes Hamkins. The axiomatic method gives us a non-judgmental platform to investigate alternatives to axioms we believe in. Early now geometers believed our space was Euclidean, but after the investigation of non-Euclidean geometries, it seems geometers are now more ambivalent. (Personally I believe AD.) – None – 2010-04-29T04:25:22.347

5

The effective/nonconstructive trade-off is also of use when going the other direction. That is, many messy epsilon management tasks in analysis can be avoided by moving the discussion to structures involving nonprincipal ultrafilters.

Or so says Tao in his exceptional blog post on the subject: http://terrytao.wordpress.com/2007/06/25/ultrafilters-nonstandard-analysis-and-epsilon-management/

– dakota – 2010-04-29T04:42:57.403

1Your second reason is the one I identify with most strongly. – Qiaochu Yuan – 2010-04-29T04:43:32.220

4Your second reason seems to me very much in the same spirit of the ones Harry and Theo gave, and that general philosophy is the one I identify with most strongly. Namely, even if AC holds in the universe of ordinary sets, there are other interesting and important places where it fails. – Mike Shulman – 2010-04-29T05:18:05.687

3Yes, Mike, I think that is right. – Joel David Hamkins – 2010-04-29T05:22:12.003

46

There are of course reasons to care about the axiom of choice because there are categories in which epimorphisms do not split. However if one sticks to the category of sets my position could be (provocatively) described as follows: The axiom of choice is obviously false but that doesn't stop me from using it.

Before I go on to explain why I think it is false let me make a general remark. Set theory is a mathematical model for mathematics it isn't mathematics itself. We all know that all models usually manage to model some part of what they model but they almost never correctly model everything. Things are a little bit more complicated in the case of set theory but it is also supposed to be the common language of mathematics. However, it really works as such only as a protocol for conflict resolution; in case we disagree over a proof we are supposed to work our way down to formal set theory where there couldn't possibly be any conflicts. However, most mathematicians would rather, I believe, voluntarily submit to extended flagellation than actually work with formal set theory. Luckily, in practice all disagreements are resolved long before one reaches that level. Furthermore, most working mathematicians show a cavalier towards set theory. It is quite common to speak of the free group on isomorphism classes of objects in some large category which is not possible in formal set theory as the the isomorphism classes are themselves proper classes and hence can not be members of some class. Of course, when pressed a mathematician using such a phrase would probably modify it by speaking of skeleta but I have once been criticised when using a slightly different formulation that avoided the problem without speaking of skeleta as being wrong for set-theoretic reasons. (This is not meant as a criticism of the person in question, a working mathematician should have the right to ignore the horny parts of set theory, at their own peril of course.)

Now, the reason why I believe that the axiom of choice is obviously false is that gives us an embedding of the field of $p$-adic numbers into $\mathbb C$ which seems fishy as they are constructed in such different ways. In fact if you try to pin down such an embedding by asking for it to fulfil more conditions then it doesn't exist. This is true if you ask that it be measurable or take definable numbers to definable numbers and so on. My own feeling is that its existence is so counter-intuitive that it couldn't possibly existst. On the other hand such an embedding is used over and over again in say the theory of $\ell$-adic cohomology. It is true that in that case at least it can be avoided (Deligne seems to share some of my disbelief as in his second paper on the Weil conjecture he starts with a short discussion on how to avoid it but still uses it as it cuts down on uninteresting arguments).

My feeling about the axiom of choice is pragmatic; it is useful and doesn't seem to get us in trouble so I have no qualms using it (even though I don't believe in it fully). I have also a picture of sets which could be used to justify this contradictory (I am not trying to formalise it so it should not be considered a competing model of mathematics). To me all elements of a set are not on equal footing. Taking my cue from algebraic geometry, there are closed points which are "real" elements but also non-closed points. Hence, the set of embeddings of the $p$-adic numbers in $\mathbb C$ is under the axiom of choice a non-empty set but in my opinion it does not contain any closed points. (In fact all its elements are probably generic, i.e., their closure is the whole set.) As long as you only deal with "concrete" objects (which should be closed in any set in which they are contained but maybe should be something more) the conclusions about them that are obtained by using the axiom of choice should be OK.

30I have to admit, this is one common misgiving I've never been able to really understand. That is, among the many non-constructive isomorphisms we encounter in mathematics, why is the one between $C$ and $\bar{Q_p}$ so odious? Isn't it supposed to be just a reflection of a sort of 'uniformity' of algebraically closed fields, as with vector spaces? I suppose the wildly different topologies give us pause, but why shouldn't a rather 'bare' structure like a field have several distinct topologies? – Minhyong Kim – 2010-04-29T13:24:44.673

4On the other hand, I'm sure you and Deligne have given much thought to this matter, while I'm very naive. Perhaps it would help my understanding if we try to refine your objection a bit. Take two sets $S$ and $T$ of the same cardinality. Do you find it unbelievable that $Q(S)$ and $Q(T)$ are isomorphic? – Minhyong Kim – 2010-04-29T13:27:15.640

2To take your question first. You are assuming we have a bijection between $S$ and $T$ which of course allows us to give a very explicit formula for the isomorphism between $Q(S)$ and $Q(T)$. This is completely constructive, there is of course a free variable or universal quantifier for the bijection but that is no problem, you get the bijection as input and don't bother where it comes from. – Torsten Ekedahl – 2010-04-29T14:18:04.910

2(cont'd) Things become more complicated when you are saying that two uncountable algebraically closed fields of the same characteristic and cardinality isomorphic. That uses the axiom of choice even with the bijection given. Note that I rather accept consequences of the axiom of choice on a case by case basis. Hence, the case of an embedding of the p-adic field into is one that I have met so many times so I had to take a stand. The case of for instance the existence of a maximal is different because there are so many natural cases where it can be proved without the axiom of choice. – Torsten Ekedahl – 2010-04-29T14:25:31.780

6I find this particular use of AC relatively harmless for the simple reason that it is very easy to get countable approximations to an isomorphism. In such cases, you can assume that the isomorphism is generic and then almost anything reasonable that you can prove using the generic isomorphism you can also prove using countable approximations instead. In other words, so long as the isomorphism is used mostly as a North Star to keep your bearings straight, you can't get into much trouble. – François G. Dorais – 2010-04-29T15:34:08.153

Torsten: So do I understand correctly that you find objectionable the statement $F\simeq K \Rightarrow \bar{F}\simeq \bar{K}$ when the objects have large cardinality? – Minhyong Kim – 2010-04-29T16:19:53.530

@François: I completely agree with you and I also tried to explain that in my original answer.

@Minhyong: What is the original isomorphism $F \simeq K$ in the case at hand? – Torsten Ekedahl – 2010-04-29T20:20:37.420

Torsten: Hmm, your question makes me worry that I'm misunderstanding something very basic. Anyways, $F$ and $K$ would be the purely transcendental extensions of $Q$ in $C$ and $\bar{Q}_p$ generated by transcendence bases. Perhaps it's the existence of transcendence bases that you don't like? – Minhyong Kim – 2010-04-29T20:46:20.317

That is, $F$ and $K$ are the $Q(S)$ and $Q(T)$ above. – Minhyong Kim – 2010-04-29T20:50:20.587

Yes, if you look at the proof of Steinitz theorem it is in the construction of transcendence bases that the axiom choice is most heavily used. (It is also used in the extension to the algebraic closure but that is much more benign.) – Torsten Ekedahl – 2010-04-30T04:12:49.483

7So when you say 'the axiom of choice is obviously false,' the most relevant statement here is 'I don't believe fields like $C$ or $\bar{Q}_p$ could have transcendence bases.' I think the reason I feel differently is because it seems plausible that there are only two ways to build fields of characteristic zero starting from $Q$: (1) Construct purely transcendental extenstions $Q(S)$; (2) Construct algebraic extensions on top of that. Since we will end up with a purely algebraic structure, it doesn't seem so strange then that many compatible topologies are possible on a given construction. – Minhyong Kim – 2010-04-30T09:03:30.770

13By the way, does it also seem 'obviously false' that $C$ and $\bar{Q}_p$ are isomorphic as sets? After all, we could be objecting even then: 'but their constructions are so different!' or 'if you try to pin down such an isomorphism by asking for it to fulfill more conditions, then it doesn't exist.' – Minhyong Kim – 2010-04-30T09:09:29.913

3There is even a measurable (defined appropriately) bijection between them so that's no problem. Let me rephrase my objection. There have been things in the past that I have considered obviously false but were wrong about. I certainly don't exclude the possibility that someone could give a convincing construction of an isomorphism which would force me to modify my intuition. However, the axiom of choice is sufficiently nebulous so that the fact that it implies the existence rather has made me put it on the restricted list of results whose acceptability I prefer to judge on a case by case basis. – Torsten Ekedahl – 2010-05-02T19:51:02.050

Torsten, can you clarify measurable? – François G. Dorais – 2010-05-02T23:46:37.817

I had in mind the $\sigma$-algebra structure on $\overline{\mathbb Q}_p$ induced by the Borel-algebra on all finite extensions of $\mathbb Q_p$ (and the standard one on $\mathbb C$). For each finite extension of $\mathbb Q_p$ there is a measurable bijection between it and $\mathbb C$. By rearranging them for different extensions one should (I haven't properly checked it though) get the bijection for $\overline{\mathbb Q}_p$. – Torsten Ekedahl – 2010-05-03T06:38:57.373

34

I would like to point out that a lot of the people who are interested in the Axiom of Choice (AC) are not worried about it in any mathematical way.

In general, I think "Why do people worry about X?" is primarily psychological. I can (and alas, do) worry about the correctness of my proofs for lots of reasons...but fundamentally I am worried about myself and not about some mathematical object or principle.

In fact, I don't see what there is to worry about concerning AC. We know it is independent of ZF, we know it has some counter-intuitive consequences, and we know that many of our most basic and fundamental "abstract" theorems either require AC to hold in full generality (in the sense that there are models of ZF in which they are false) or in fact are equivalent to AC: e.g. every vector space has a basis (equivalent), the product of quasi-compact spaces is quasi-compact (equivalent), every field has an algebraic closure (required), every ring has a maximal ideal (equivalent),...I think we have all the information that we need on the role of AC in mathematics.

On the other hand, if you encounter a theorem, it is often an interesting question to ask whether the proof uses AC and, if it does, whether it can be modified so as not to require AC or only to require some weaker form of AC. As an example, I asked the following MO question

How much choice is needed to show that formally real fields can be ordered?

and the answer was a reference to a paper where it was shown that the orderability of formally real fields is equivalent to the Boolean Prime Ideal Theorem. This is an interesting paper; I learned more about field theory (not set theory) by reading it.

Why is it interesting to ask whether a proof requires AC? For starters, for good reasons (not worries!) people are interested in making proofs constructive as much as possible. You say that you've proven that for every number field $K$, and positive integer $n$, there exists a genus one curve $C_{/K}$ such that the least degree of a closed point on $C$ is equal to $n$? (Yes, I have.) OK, I'll give you $K$ and $n$: can you give me an explicit set of defining equations defining such a curve? (No, I can't!) My theorem would be better if it could be made explicit -- it's obvious, isn't it?

It is a generally believed "meta-theorem" that if a result can be shown to require AC, then it is certifiably non-constructive. Thus, if one is trying to make a certain result constructive, one would certainly like to know whether AC is essentially involved in its proof.

32

The way I always justified the usage of the axiom of choice to myself (which I haven't had to do very often) is as an expositional convenience. If somebody told me that I couldn't use the axiom of choice anymore, I would take all the theorems I know about commutative rings and replace "Let R be a commutative ring" with "Let R be a commutative ring with a maximal ideal", and then I would go about my business, since nobody cares about the crappy ring you slapped together which doesn't have a maximal ideal.

1I pretty much share your attitude. I guess I asked this question to figure out if this attitude costs me anything, because apparently there are people who REALLY keep track of this stuff... – Paul Siegel – 2010-04-29T03:37:29.577

19There's a better alternative: prove that every non-crappy ring has a maximal ideal. Logicians have been in that business for a very, very long time. Like all mathematicians, logicians like a good counterexample and, contrary to common belief, that's the real reason why logicians slap together some really crappy rings from time to time. – François G. Dorais – 2010-04-29T03:42:26.777

3@Paul: People who really keep track of that usually have a good reason to do so, which is not necessarily relevant to working mathematicians using the results. – François G. Dorais – 2010-04-29T03:55:03.927

10

Let me suggest this lovely paper by Thierry Coquand and Henri Lombardi: "A Logical Approach to Abstract Algebra" (http://en.scientificcommons.org/42279746). It uses Barr's theorem (any geometric consequence of a geometric theory can be derived intuitionistically, if it follows from classical logic+choice) to turn some "ideal-istic" results into much simpler linear algebra.

– Neel Krishnaswami – 2010-04-29T12:53:14.000

1Well, I don't need full choice to construct nontrivial ultrafilters on $\mathbb N$, but I do need some weaker version. Ultrafilters are precisely the maximal ideals in some ring, which is decidedly something worth caring about, whether or not it is crappy. – Theo Johnson-Freyd – 2010-05-01T17:35:20.180

24

When I first encountered AC as an undergraduate student, like most math students I thought it was seriously questionable since it led to weird and counter-intuitive things like non-measurable sets and, worst of all, the Banach-Tarski Paradox.

But after I learned that AC is logically equivalent to

 **The cartesian product of any non-empty collection of non-empty sets is non-empty**


I found it impossible not to believe it. My only conclusion could be that my mathematical intuition was not well-developed.

I came to accept the consequences of AC as natural aspects of mathematics, and they no longer seem nearly as weird or counter-intuitive as they did at first.

And I would be fairly unhappy if there existed a vector space without a basis. Or if there existed two sets A, B without an injection, bijection, or surjection A -> B (i.e., a failure of Trichotomy).

On the other hand, I have begun to be philosophically troubled by the common and probably harmless assumption that mathematicians can choose between the two complex roots of

(*)       z2 + 1 = 0

(and similar situations). There is no basis for selecting between i and -i . . . or even naming them i and -i ! So although I intend to continue referring to i for convenience, it feels to me that technically one has no right to do so; instead a technically correct discussion should always refer to the set of roots of (*), but never just one of them alone.

1Wait, the one that's above the real line is $i$, and the one below is $-i$, right? ;) – Vladimir Reshetnikov – 2013-08-16T01:26:19.823

6If you construct $\mathbb{C}$ as $\mathbb{R}^2$ with product $(a,b)\cdot (c,d):=(ac-bd,bc+ad)$, then $i:=(0,1)$ is a standard definition. Also if you construct $\mathbb{C}$ as $\mathbb{R}[x]/(x^2+1)$ you have a standard choice: $i:=x\mathrm{mod}(x^2+1)$. – Qfwfq – 2013-10-08T23:13:12.410

3Another fine equivalence to AC: every set has a unique cardinality. This and your first example are the main things that convince me AC is not so strange. – Neil Toronto – 2013-10-10T14:59:31.833

8@NeilToronto: “every set has a unique cardinality” can be phrased in a lot of different ways, plenty of which are provable without choice (e.g.: there is a class $\mathbf{Card}$ and a “cardinality” map $V \to \mathbf{Card}$, such that sets have the same cardinality iff they are isomorphic). The only versions of the statement I know that are equivalent to AC are ones which insist that each cardinality should be represented by some ordinal — but this is a very thinly veiled version of the well-ordering principle, and I think not at all so intuitively obvious. – Peter LeFanu Lumsdaine – 2013-10-10T17:21:29.220

The two statements are very close to each other so it might worth to explore why we have mentally different intuitions about the two statements. I think it might have to do with the word "choice" as if we need some action performed by an agent while for the other one there is no such word. – Kaveh – 2013-10-16T01:32:22.823

It isn't natural to believe that a product of nonempty sets is nonempty once you generalize a bit: an $I$-indexed family of sets ${ Ji }{i\in I}$ is the same as an epimorphism $J \to I$, $J = \sum_{i\in I} J_i$. An element of a product of this family is the same as a section of this epimorphism --- and of course epimorphisms in categories can have no sections! This is true even in categories that are a model of (extensional) set theory, i.e. in toposes like a category of sheaves of sets on a space. E.g. for any nontrivial manifold $X$ its $Sh(X)$ will not satisfy AC. – Anton Fetisov – 2018-01-23T18:22:22.760

16

I'll give you a reason that applies to people who aren't just philosophically prejudiced:

Anything proven without resorting to the law of excluded middle (which is implied by the axiom of choice) is true in any elementary topos. There are certainly people who don't like AC for philosophical reasons, but this is a good reason for any mathematician of any persuasion.

1The law of excluded middle is "implied by the axiom of choice" ?? – Ewan Delanoy – 2010-04-29T03:05:28.807

16Well, at most you've reduced the question to the definitely less sexy «Why worry about statements true in any elementary topos?»! – Mariano Suárez-Álvarez – 2010-04-29T03:06:41.310

2@Ewan: Yes, I don't know how hard the proof is, but I think that it was proven in the 1960s or 1970s, but you should probably ask François or Joel if you want more details. – Harry Gindi – 2010-04-29T03:08:01.617

@Mariano: Well, in particular, the results hold for all grothendieck toposes as well, which are not in general boolean. – Harry Gindi – 2010-04-29T03:09:27.683

6That's Diaconescu's Theorem: every topos which satisfies the Internal Axiom of Choice (IAC) is Boolean. – François G. Dorais – 2010-04-29T03:17:46.607

7@Mariano: It might be less sexy, but it's easier to answer, e.g. statements true in the effective topos are effectively true. – François G. Dorais – 2010-04-29T03:35:18.600

12Yeah, but you can keep the axiom of choice if you surrender the powerset axiom. (E.g., in Martin Lof's constructive type theory, choice is actually a theorem.) It's the combination of impredicative sets and choice that is funny (at least from a constructive pov).

Another way of putting it is that internal logics of categories offer reasons to keep track of every combination of axioms, even superbasic stuff like contraction and weakening -- choice is not exceptional in this regard. – Neel Krishnaswami – 2010-04-29T10:39:04.660

16

If you are doing analysis, presumably you want to use Lebesgue measure. In order for Lebesgue measure to have its usual properties, it is essential that $\mathbb{R}$ is not a countable union of countable sets.

14

It is a mistake to think that the axiom of choice has no relevance to, say, undergraduate mathematics. The axiom of choice makes undergraduate analysis easier by enabling one to say that $f(x)$ is continuous at $x=a$ if $f(a_n)$ tends to $f(a)$ for each sequence $\langle a_n \rangle$ that tends to $a$.

The popular undergraduate text Understanding Analysis by Stephen Abbott owes its success (at least partly) to the use of this definition rather than the $\varepsilon-\delta$ definition. But the definitions are equivalent only assuming some form of the axiom of choice. (The book does not mention this, however.)

9Actually, it is provable in ZF (without any form of the axiom of choice) that a function from R into R is continuous if and only if it is sequentially continuous. (Theorem 4.52 of "Axiom of Choice" of Herrlich).

Most of the real analysis, contrarily to popular belief, is still in force without any form of the axiom of choice. – Steven – 2010-10-21T16:30:26.917

20I'm not a set theorist (I should really make a macro for this prefatory invocation), but: isn't it only an extremely weak version of AC that is needed here? Something like countable choice? As Kaplansky says in his Set Theory and Metric Spaces, really countable choice should go without comment in any place where non-axiomatic set theory is being discussed. I think that 99.9% of practicing mathematicians would regard countable choice as simply being "true". The fact that it is not formally implied by ZF is not a strike against it, but rather against ZF itself. – Pete L. Clark – 2010-04-29T06:00:08.880

5Jonhn's point may be expressed as "sequential continuity of $f$ implies pointwise continuity of $f$". This has very little to do with choice. As Pete observes, countable choice is surely sufficient. In constructive mathematics (with countable choice) the implication "sequential continuity implies pointwise continuity" is equivalent to Hajime Ishihara's BD-N principle which states: "A set $A \subseteq \mathbb{N}$ is bounded if, for every sequence $(a_n)_n$ in $A$, $\lim_n a_n/n = 0$". The principle holds classically as well as in many intuitionistic models. – Andrej Bauer – 2010-04-29T06:48:50.470

2It's true that only a weak form of AC is needed to prove that sequential continuity implies continuity, such as countable choice. You get a counterexample to this implication in a model of ZF with a Dedekind finite set of reals. My point is that even a little bit of AC makes a big difference to proofs in undergraduate analysis. – John Stillwell – 2010-04-29T08:08:48.577

4Unfortunately, there are still many topoi in which countable choice fails, so those of us who want to avoid choice in order to make our proofs applicable in any topos still have to avoid countable choice. – Mike Shulman – 2010-05-03T14:03:14.970

@Steven But John's answer is about continuity at a point, which doesn't hold in $\mathbf{ZF}$, and not about the "global continuity" of a real function (which is related, but different), which does hold in $\mathbf{ZF}$. – Maxis Jaisi – 2018-01-09T07:19:14.520

10

Many years ago I have been a professor for abstract analysis and had no reservations whatever against using the axiom of choice or "equivalent" statements as e.g. Zorn's lemma since many results in functional analysis depend heavily on it. I have also been very impressed by nonstandard analysis especially in the version of Nelson. But in the mean time I have become more skeptical. I am still very impressed by the results mathematics has obtained by treating the infinite world analogous to the finite world, but I have the feeling that there are some sorts of levels in mathematics which should not be confused. This is of course a very vague assertion. But consider the general statement that every vector space has a basis. This belongs to the infinite world which is "far away". Here Zorn's lemma seems to be the appropriate means. Then consider the statement that each continuous solution of the functional equation f(x+y)=f(x)+f(y) over the reals has the form f(x)=ax. This belongs to the finite world. But the statement that there exist discontinuous solutions mixes both worlds. It uses the existence of a Hamelbasis for the reals over the rationals. Here I am not sure what this really means. It seems that this gives an explanation of what discontinuous solutions look like. But in fact it gives only an illusion of what such solutions look like.

6

As Pete Clark says,

"In general, I think "Why do people worry about X?" is primarily psychological. I can (and alas, do) worry about the correctness of my proofs for lots of reasons...but fundamentally I am worried about myself and not about some mathematical object or principle."

Let me explain why I worry about the axiom of choice and, generally, about non-constructive methods in mathematics. One of the fascinating aspects of mathematical activity is that it produces results that can transcend our life, like the Pythagorean theorem or Newton's binomial formula.

Many concepts of (formalized) mathematics evolved as ideal models of concrete things: some functions can model the movement of a particle, Turing machines model something following an algorithm, etc.. Although the relationship between the abstract, formal concept and the "real" thing is sometimes loose, it is a historical fact that the concepts were devised as models of such concrete things.

Mathematical theorems proved in a constructive way generally have a close connection to algorithmic processes that can possibly be relevant for humans of all times. But for example, the axiom of choice establishes a similarity between finite and infinite sets (which are qualitatively different, for example, no finite set is bijective to a proper subset) that is difficult to justify from an objective point of view. It states that infinite products of nonempty sets are nonempty. As far as I see, there isn't an objective argument to prefer the axiom of choice to it's negation. Future generations of mathematicians could possibly negate it (or more probably, live it undecidable) and a lot of parts of twentieth century mathematics would cease being theorems.

However, it seems less likely that a future mathematical language would cease providing concepts that mimick, for example, automata.

Note: The undecidability of the axiom of choice in ZF theory is still a metatheorem, so if a ZF statement can be proved false using AC, then it is (meta) clearly not a theorem. So the axiom of choice could still be useful for studying ZF, as a tool to produce counterexamples that limit our theorem-proving pretensions, but with the same formal status as, say, its negation.

Also note: I see that this answer is not part of math if math is understood as the study of the consequences of the ZFC axioms, and also it doesn't deal with the specific questions, but I hope that it explains why (some) people worry and fuss about the axiom of choice. It may be relevant, for example, for math professors, that could atract a wider audience by presenting a more secular version of mathematics that excludes the axiom of choice from the set-theoretic formalism. Independently of whether they use it or not in their own work.

5

We have a proof that AC is independent of the other axioms of ZF. But that's not a good enough criterion for deciding whether or not to use an axiom. We choose axioms because because we want to use them to reason reliably. We can use ZF to reason about mundane things like strings of symbols encoded with Godel numbering. But Godel's Incompleteness Theorem shows that we can find axioms independent of ZF, which when added to ZF, give incorrect results when applied to strings of symbols. I think some people worry that AC may allow us to reason incorrectly in some domains, even if it is consistent with ZF. (Of course we don't know ZF itself is consistent, or allows us to reason correctly, either, but the fewer axioms the better.)

9Moreover, it's standard practice for mathematicians to question the necessity of their hypotheses. – François G. Dorais – 2010-04-29T04:03:04.363

4But it is far less standard to question one's axioms, after they are indentified as such. – Joel David Hamkins – 2010-04-29T05:01:33.823

12The only difference I can think of between an axiom and a hypothesis is that an axiom is a hypothesis that lots of people tend to assume. – Mike Shulman – 2010-04-29T05:12:24.193

Fair point, but there are other instances, e.g. non-Euclidean geometry. – François G. Dorais – 2010-04-29T05:14:52.240

4

Daniel Asimov's answer regarding the multiplicative axiom (Russell's terminology) seems important, as it is very difficult to disbelieve that statement (on cardinality principles alone). But I think the key is that AC is equivalent to the well ordering theorem, which is easy to disbelieve and untrue in some models of ZF. And of course, infinite cardinality requires AC to get a well-defined order, so the intuitiveness of the multiplicative axiom is less intuitive than it seems.

I like a comment James Munkres made in his Topology textbook regarding the construction of the first uncountable ordinal: you can do it without AC, but then you can't use it for anything! It's that property that you can DO something with your choice that produces both beauty and monsters, and I think this is intrinsically related to some kind of well-ordering.

I really like Horst Herrlich's Springer Lecture Note on AC (#1876), which is both short and comprehensive about what choice does across multiple disciplines. While it thoroughly discusses the well-known pathologies ("Disasters", as he formally names them!), it also discusses more fundamental issues like how the various equivalent notions of compactness fly apart without AC.

2

Another useful example is the statement that $M$ and $M^2$ have the same power for any infinite set $M$. This is proved with the use of the axiom of choice. This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the $l_1$-basis in $G$ has the same power as $M$. (See paper of mine in: Siegel W., Duplij S. and Bagger J., eds. (2004), Concise Encyclopedia of Supersymmetry And Noncommutative Structures in Mathematics and Physics, Berlin, New York: Springer-Verlag; math-ph/0009006.) To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the carpet'') need the axiom of choice.