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The following is not a proper mathematical question but more of a metamathematical one. I hope it is nonetheless appropriate for this site.

One of the non-obvious consequences of the axiom of choice is the Banach-Tarski paradox and thus the existence of non-measurable sets.

On the other hand, there seem to be models of Zermelo-Fraenkel set theory without axiom of choice where every set would be measurable.

What does this say about the "plausibility" of the axiom of choice? Are there reasons why it is plausible (for physicists, philosophers, mathematicians) to believe that not all sets should be measurable? Is the Banach-Tarski paradox one more reason why one should "believe" in the axiom of choice, or is it on the opposite shedding doubt on it?

ThiKu
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    Personally, I use one side of my brain to think about physics and a second side to think about set theory. I don't think there could possibly be any physical incarnation of the Banach Tarski paradox, or physical intuition about it, or even geometric intuition. If it did exist in the physical world, It'd probably be such an anomaly that'd become the next biggest thing to study like the Higgs Boson. But that's just my two illiterate cents. –  Jan 20 '17 at 02:16
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    Banach Tarski rules out finitely additive measures that measures every subset of $\mathbb{R}^3$. In $\mathbb{R}^2$ and $\mathbb{R}^1$ such measures exist. If your quarrel is with the idea of non-measurable sets, the usual argument (not the Banach Tarski one) which shows that there are no countably additive, translation invariant measures on $\mathbb{R}$ that measures every set (assuming the axiom of choice) seems a much better starting place. – Willie Wong Jan 20 '17 at 03:02
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    @WillieWong By the way, the idea behind both is due to Hausdorff (essentially, Banach and Tarski only popularized it.) – Anton Petrunin Jan 20 '17 at 03:26
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    @AntonPetrunin: excellent point! Indeed, if one looks at the proofs of the Banach-Tarski paradox and of the nonmeasurable set in $\mathbb{R}$, both reduce to a first step of analyzing the symmetry group of the space and a second step of using this group structure + axiom of choice to get nonmeasurable sets. In some sense the striking part about Banach-Tarski (that finitely many pieces suffice) has more to do with geometry than to do with axiom of choice. This is sort of what I was referring to in my previous comment. – Willie Wong Jan 20 '17 at 04:37
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    @WillieWong actually, in the same paper Hausdorff proves a vesion of "Banach-Tarski paradox" for sphere whithout countable set of points (from this the standard vesion follows easily). So I think it is better to name it after Hausdorff. – Anton Petrunin Jan 20 '17 at 05:05
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    Seconding what James Nixon said: to my mind, Banach-Tarski is a theorem about mathematical objects, not physical ones. That those mathematical objects happen to be bounded sets in $\mathbb{R}^3$ and that we can also physically conceive of some bounded sets in $\mathbb{R}^3$, does not mean that these particular sets have any relevant physical interpretation; and that lack of interpretation is (again, to me) completely independent of AC. – Steven Stadnicki Jan 20 '17 at 05:26
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    @WillieWong You mean, they rule out finitely additive measures on $\mathbb R^3$ which are nontrivial and invariant under rigid motions. – bof Jan 20 '17 at 10:26
  • @AntonPetrunin Did Hausdorff also prove the more general version of the paradox which says that two bounded sets with nonempty interiors (e.g. balls of different size) are equivalent by finite decomposition? Deducing the generalization from the equivalence of one ball to two balls is elementary but not quite trivial, I think. – bof Jan 20 '17 at 10:29
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    No mathematical model of physical phenomena is perfect: they all break down when the assumptions of the model aren't satisfied. Quantum mechanics already suggests that real vector spaces are imperfect models of physical space. Why be surprised that even finer-grained phenomena (such as the Banach--Tarski paradox) don't carry over? – HJRW Jan 20 '17 at 11:30
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    I would turn the first sentence around -- non-measurable sets exist, and consequently you have things like the B-T paradox. BT isn't all that surprising once you realize all its saying is "No, really, measure doesn't work well when non-measurable sets are involved, not even if you start and end with measurable ones". –  Jan 20 '17 at 12:07
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    Reminds me of the famous Jerry L. Bona quote: “The Axiom of Choice is obviously true, the Well–ordering theorem is obviously false; and who can tell about Zorn’s Lemma?" – Drunix Jan 20 '17 at 14:29
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    @HJRW: Exactly right. The axiom of choice requires not just fine-grain but uncountably infinitely fine-grain! – user21820 Jan 20 '17 at 15:32
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    Somewhat related: http://mathoverflow.net/questions/238153/physical-meaning-of-the-lebesgue-measure – Terry Tao Jan 20 '17 at 16:50
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    I'm a physicist, and I don't believe in the real number system. You can't measure the distinction between a rational number and an irrational number, so the distinction is physically meaningless. But the reals are convenient. If I wanted to do all my physics using only ultrafinitist methods, I could in principle do that, but it would be inconvenient and I would have to learn a lot of specialized math. Physicists know how to jump into an artificial mathematical world, do a calculation, and then jump back out and translate the result into real-world predictions. –  Jan 20 '17 at 17:56
  • Many years ago we had an undergrad in our department design a t-shirt that read "I believe the Axiom of Choice" on the front and had a symbolic depiction of the axiom on the back. I bought one of the shirts and wore it proudly until it wore out! – Jim Conant Jan 20 '17 at 20:23
  • @Drunix: I've always found that quote weird; I find the well-ordering theorem just as obviously true as choice. And both are for essentially the same reason: proof by transfinite iteration. –  Jan 20 '17 at 22:39
  • I think of it as follows: ​ The Axiom of Choice is magic, and the Axiom of Determinacy is physics. ​ ​ ​ ​ –  Jan 22 '17 at 13:16
  • 'not all sets should be measurable' --> not all sets should be Lebesgue measurable ? i mean it's easy to construct measures where not all sets are measurable without using axiom of choice right? – BCLC Sep 14 '21 at 04:36
  • Complex numbers in physics are both real (quantum states are inherently complex) and unreal (you can't measure them). It's entirely possible unmeasurable sets are the same. You can't create an unmeasurable chunk of a ball, but maybe dark energy is best modeled by a quantum Banach-Tarski process due to (I'm making this up) renormalization space being a nonmeasurable set or something. I think the question is not "is it real", but "is it useful". (From there you could go down the "maybe nothing is real, and useful is all there is" path). – Dax Fohl Jun 20 '23 at 01:17

9 Answers9

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There are two ingredients in the Banach-Tarski decomposition theorem:

  1. The notion of space, together with derived notions of part and decomposition.
  2. The axiom of choice.

Most discussion about the theorem revolve around the axiom of choice. I would like to point out that the notion of space can be put under scrutiny as well.

The Banach-Tarski decomposition of the sphere produces non-measurable parts of the sphere. If we restrict the notion of "part" to "measurable subset" the theorem disappears. For instance, if we move over into a model of set theory (without choice) in which all sets are measurable, we will have no Banach-Tarski. This is all well known.

Somewhat amazingly, we can make the Banach-Tarski decomposition go away by extending the notion of subspace, and keep choice too. Alex Simpson in Measure, Randomness and Sublocales (Annals of Pure and Applied Logic, 163(11), pp. 1642-1659, 2012) shows that this is achieved by generalizing the notion of topological space to that of locale. He explains it thus:

"The different pieces in the partitions defined by Vitali and by Banach and Tarski are deeply intertangled with each other. According to our notion of “part”, two such intertangled pieces are not disjoint from each other, so additivity does not apply. An intuitive explanation for the failure of disjointness is that, although two such pieces share no point in common, they nevertheless overlap on the topological “glue” that bonds neighbouring points in $\mathbb{R}^n$ together."

Peter Johnstone explained in The point of pointless topology why locales have mathematical significance that goes far beyond fixing a strange theorem about decomposition of the sphere. Why isn't everyone using locales? I do not know, I think it is purely a historic accident. At some point in the 20th century mathematicians collectively lost the idea that there is more to space than just its points.

I personally prefer to blame the trouble on the notion of space, rather than the axiom of choice. As far as possible, geometric problems should be the business of geometry, not logic or set theory. Mathematicians are used to operating with various kinds of spaces (in geometry, in analysis, in topology, in algebraic geometry, in computation, etc.) and so it seems only natural that one should worry about using the correct notion of space first, and about underlying foundational principles later. Good math is immune to changes in foundations.

Andrej Bauer
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    I think the reason everyone isn't using locales is that for most intents and purposes, topology is enough. For everyone to switch from a simple concept to a more complicated one, we would need to have significant (i.e. interesting and known to many people) applications. – tomasz Jan 20 '17 at 13:25
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    I second this answer as much as I can. As you mention, the "issue" with the notion of a space really can be seen as the one which causes the clash with physical intuition. After all, a layman confronted with Banach-Tarski paradox explained using apples, gold balls or whatever, will most likely first argue by "but in reality you can't cut it like that [because atoms or whatever]". – Wojowu Jan 20 '17 at 16:36
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    @tomasz : I see an analogy with the Riemann integral vs. the Lebesgue integral. For (most) scientists and engineers, the Riemann integral is enough, and they are not going to be motivated to study the seemingly more "complicated" Lebesgue integral, even though the Lebesgue integral has cleaner and more satisfying theoretical properties. – Timothy Chow Jan 20 '17 at 19:18
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    @TimothyChow: the analogy is not complete. The definition of Lebesgue integral is not really any more complicated than that of the Riemann integral, the complicated part is defining the Lebesgue measure. Besides, mathematicians in general do use Lebesgue measure. The reasons to use Riemann integral are mostly inertia and relative ease of calculation (as far as I know). – tomasz Jan 20 '17 at 20:25
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    I find the statement that locales are more complicated than topological spaces similar to stating that (abstract) groups are more complicated than matrix groups. A topological space can be defined as a set together with a collection of its subsets with certain properties. A locale can be viewed as (an abstractly given) second part without the first one. That is, a structure with properties that open subsets of a topological space must satisfy, without the requirement that elements of this structure are subsets of something. It is thus more abstract, but is it more complicated because of that? – მამუკა ჯიბლაძე Jan 20 '17 at 20:49
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    @tomasz : Your comments actually confirm the analogy for me. I am not convinced that locales are more complicated than topological spaces. And I also think that resistance to locales is largely inertial. Finally, there is a sizable community of people who use locales. – Timothy Chow Jan 20 '17 at 21:10
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    @მამუკაჯიბლაძე: Fair enough. I admit I am not familiar with locales, and the linked definition is not very welcoming, but that's probably just the way nlab articles tend to be. – tomasz Jan 20 '17 at 21:34
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    @tomasz: Peter Johstone's paper might be a bit more illuminating. – Andrej Bauer Jan 20 '17 at 22:17
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    @AndrejBauer: Yes. As a matter of fact, I'm reading it right now and it's pretty interesting. :) – tomasz Jan 20 '17 at 23:22
  • @AndrejBauer: I have read Johnstone's survey, too, and I'm ready to start believing in the strength of locales. Can you mention a concrete example which can be more efficiently topologized by locales (rather than by point topologies)? – Delio Mugnolo Jan 30 '17 at 10:19
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    For me personally it is not so much a matter of "belief" but rather staying open to many points of view and readiness to adopt new concepts when they have advantages. As for your question, I can think of two: Alex Simpson's explanation of the space of random sequences seems natural. Another, better known, is Stone duality (of any kind) when we make a spectral space (of a Boolean algebra, or a good enough ring, etc.). The spectrum of a space is most naturally described directly in terms of its topology: a basis for it are the elements of the algebraic structure. – Andrej Bauer Jan 30 '17 at 10:23
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    Another, technically more relevant example is how we think of spaces in the context of computability theory and computation. A basic observation about a real is that it belongs to some interval $(p,q)$. A real is a certain collection of such observations (a completely prime filter), but the important thing is that the observations generate a locale which is directly useful for computation and various algorithms. That is, real-valued function do not map reals to reals, instead they transform basic observations, and that's a localic point of view. – Andrej Bauer Jan 30 '17 at 10:26
  • Thanks, rather interesting. It sounds like pursuing this philosophy might simplify standard measure theoretical statements based on the "a.e."-caveat.

    That said: Is there any metrizable space that allows for an interesting pointless topology? Fréchet spaces even?

    – Delio Mugnolo Jan 30 '17 at 10:43
  • Every space is also a locale, but I suppose you're asking if there are locales without points that are metrizable? What is the question? – Andrej Bauer Jan 30 '17 at 21:17
  • @AndrejBauer Your comment mentioning the localic and computable point of view on real numbers has me intrigued. Do you happen to have a reference that develops this further? – ಠ_ಠ Dec 02 '17 at 04:49
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    I'd say you should read @PaulTaylor 's work on analysis (there are a number of references on that page, but maybe start with Foundations for computable topology). If that is too radical for you, come back and we'll find something more mundane. – Andrej Bauer Dec 02 '17 at 22:23
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It's notable that most of the "bread and butter" mathematical consequences of the axiom of choice are actually consequences of countable choice. (Every infinite set contains a countable subset, a countable union of countable sets is countable, etc.) The Hahn-Banach theorem is a counterexample, but only if you want it for nonseparable spaces, and I can't think of any time I've ever needed this. When restricted to separable Banach spaces it doesn't require any choice principle at all! Whereas the seemingly pathological consequences of choice (existence of nonmeasurable sets, Banach-Tarski, well-ordering of the real line) generally do not follow from countable choice.

So the argument from mathematical value seems to me to support countable choice more than full choice. But that isn't a very strong argument, is it? We can't decide whether an axiom is true based on whether we like its consequences. At best it's suggestive.

Incidentally, I had the impression when I read Zermelo that he had great polemical skill, but none of his arguments seemed to get directly to the truth of the axiom. He argues for the mathematical value of the axiom. He points out that his critics have themselves on occasion unwittingly used the axiom, which is a devastating point, but has little bearing on the question of truth. (If I'm not mistaken, those unwitting uses were all of countable choice, by the way.)

You ask if one should "believe" the axiom of choice, and I think you are right to put the word "believe" in quotes. I feel strongly that set-theoretic assertions are objectively meaningful, but I also feel that philosophers of mathematics have done a very poor job of clarifying what sets are. (Halmos: "A pack of wolves, a bunch of grapes, or a flock of pigeons are all examples of sets of things." Black: "It ought then to make sense, at least sometimes, to speak of being pursued by a set, or eating a set, or putting a set to flight.") If we can't even get that straight, it's hard to come to grips with questions about the truth of questionable axioms.

Nik Weaver
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    Another very important consequence of AC is that, in a commutative ring with $1$, every proper ideal is contained in a maximal ideal. It seems to me that this is not a consequence of countable choice, at least for non-noetherian rings. – Francesco Polizzi Jan 20 '17 at 14:58
  • Good point, although for countable rings you don't need any choice for this result. – Nik Weaver Jan 20 '17 at 15:40
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    The uncountable rings that appear in analysis typically come with natural topologies with respect to which they are separable, and one can use this. If you're talking about untopologized uncountable rings, I wouldn't consider it "bread and butter" mathematics (just my opinion, of course). – Nik Weaver Jan 20 '17 at 15:42
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    I'm curious about your last paragraph. Why is it that you feel that set-theoretic assertions are objectively meaningful, in the sense of having ontological status beyond the syntactical formal system? Specifically for example, does the axiom of comprehension have objective meaning in the way it does not allow constructing the universal complement of a previously constructed set $S$ even though membership in the complement is simply $\neg x \in S$? – user21820 Jan 20 '17 at 15:44
  • @user21820: I think you're using the word "meaning" in a different way than I am. I agree that comprehension seems somewhat ad hoc, but on the other hand it does have a clear constructive justification. (If we can decide, for each $x \in S$, whether $x$ has property $P$, then we can construct the desired subset by hand, at least on my interpretation of these terms.) – Nik Weaver Jan 20 '17 at 15:51
  • @NikWeaver: Hmm do you mind explaining what kind of "meaning" you have in mind then? As for the comprehension axiom, your justification seems to apply equally to the universal complement, since if $S$ has been previously constructed we can 'construct its complement using it'. – user21820 Jan 20 '17 at 15:53
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    As for eating... We can eat a set of chocolate chess pieces. =) – user21820 Jan 20 '17 at 15:57
  • "your justification seems to apply equally to the universal complement" --- no, because I don't know how to construct the universe. If a set is given constructively this means I know how to construct it. For any decidable property, I can follow the same procedure but only accept the elements which have the property. – Nik Weaver Jan 20 '17 at 16:24
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    Although not very relevant to the discussion at hand, I'd like to point out that a very important and widely applicable result, namely the ultrafilter lemma, depends on more than just the countable (or even dependent) choice. Indeed, there are models of ZF in which DC holds but there are no nonprincipal ultrafilters on any set. – Wojowu Jan 20 '17 at 16:27
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    "We can't decide whether an axiom is true based on whether we like its consequences" - Well.... I would say whether we like or not its consequences is an important point for deciding if we want to pursue an axiom! – Qfwfq Jan 20 '17 at 20:34
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    @user21820 what are the elements of that set? What if I eat half of a rook? :-) – David Roberts Jan 21 '17 at 00:43
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    "because I don't know how to construct the universe." -- That's one possible justification, but then it rejects the power-set axiom (you can't construct almost all subsets of the natural numbers). My point is that I can readily accept the validity of various justifications for meaning of each ZF axiom schema, but none of them work for all the schemas, and are incompatible if we attempt to combine them, at least from my viewpoint. – user21820 Jan 21 '17 at 02:03
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    @Wojowu: But is there any theorem that over ZF requires the ultrafilter lemma to prove and has any real-world significance? – user21820 Jan 21 '17 at 02:06
  • @DavidRoberts: Eating is an operation, and if you chew off half a rook then you would have changed ${rook}$ into ${half(rook)}$. =) – user21820 Jan 21 '17 at 02:58
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    @user21820 Ultrafilters can be used to give rather neat proofs of many combinatorial statements, like Ramsey or Hindman theorem. Whether you count it as a "real-world" application might be arguable, but they are definitely quite down-to-earth applications. – Wojowu Jan 21 '17 at 05:05
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    @user21820: indeed, I do reject the power set axiom. I've written extensively about this; there are a bunch of papers on the arXiv, if you're interested. – Nik Weaver Jan 21 '17 at 06:09
  • @NikWeaver: Yes I'm very interested. I've just looked at your pdf linked from http://mathoverflow.net/a/133598/50073 and much of it is similar to my personal viewpoint, except that I think it makes more sense to have a universal type and Kleene's 3-valued logic instead of an open domain supported by intuitionistic logic. And do you agree with this post that the iterative conception of the cumulative hierarchy does not explain Replacement? – user21820 Jan 21 '17 at 15:43
  • That's hard to answer because I don't find the iterative conception cogent. – Nik Weaver Jan 22 '17 at 01:09
  • @NikWeaver: In that post I was assuming that the natural numbers and power-set operation was granted, and that one could get the union of the levels labelled by any sequence. I don't see a serious conceptual problem if you interpret the power-set operation as obtaining just the definable subsets. But even then one can never get an uncountable number of levels, since there's no sequence of uncountable length to begin with. Anyway do you think intuitionistic logic is preferable to a 3-valued logic for impredicative collections, and why? – user21820 Jan 22 '17 at 15:19
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    It's hard to answer that in a comment ... can I refer you to my book Truth & Assertibility, Chapters 4 and 6? – Nik Weaver Jan 22 '17 at 16:25
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    @NikWeaver "We can't decide whether an axiom is true based on whether we like its consequences". But isn't that how axioms where introduced? People didn't prove theorems from set-theoretic axioms before the 20th century, and when the axioms were invented, weren't they designed in such a way that they would express all that people thought was true mathematics as true theorems? – Stefan Rigger Jan 22 '17 at 17:32
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    @StefanRIgger: Yes, I think the ZF axioms were formulated in a very ad hoc way, and the nonsensical "iterative conception" was only cooked up later. Do you dispute my assertion? You feel we can decide whether an axiom is true based on whether we like its consequences? – Nik Weaver Jan 23 '17 at 00:20
  • @NikWeaver I wouldn't put it that way, but it's somewhere along the lines of what I always thought (I don't have expert knowledge of set theory, so that might very well be far from the truth). How do you judge if an axiom is true if not by its consequences? – Stefan Rigger Jan 23 '17 at 14:33
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    he Hahn-Banach theorem is a counterexample, but only if you want it for nonseparable spaces, and I can't think of any time I've ever needed this.

    The existence of Banach limits is an obvious case where HB is needed in its non-separable version. Worse even, the space of bounded linear operators on any infinite dimensional Hilbert space is not separable, so most operator theory would fail without non-separable HB.

    – Delio Mugnolo Jan 30 '17 at 10:27
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    @DelioMugnolo: I've been working with $B(H)$ for twenty-five years and to the best of my recollection I've never needed to use Hahn-Banach on it. "Most operator theory would fail" --- can you give a specific example? – Nik Weaver Jan 30 '17 at 16:36
  • @NikWeaver Well, even without dealing with $B(H)$, Hahn-Banach is used to prove basic properties of operator semigroups, like the existence of duality sets needed in the Lumer-Phillips Theorem. If one studies so-called open quantum systems, operator semigroups acting on the space $\mathcal S_1$ of trace class operators must be introduced. Even at a much more elementary level, the whole theory of $H$-valued integration and $H$-valued holomorphic functions need HB. – Delio Mugnolo Jan 30 '17 at 20:39
  • Questionable axioms? There is no doubt that the axiom of choice has been proven wrong. Its consequence, every set can be well-ordered, is as "real" as the result: there are ten different natural numbers less than 3. –  Aug 05 '17 at 15:12
  • It may be worth mentioning that countable choice is implied by a large number of other axioms, some of which are mutually incompatible with choice, and that it is constructively valid. Also, countable choice is arguably necessary for finite sets to really be well behaved with an injection from the natural numbers to any infinite set. Without it, you can have an infinite set that cannot be written as the union of two nonempty infinite sets. – saolof Sep 19 '22 at 08:20
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The other answers don't seem to have said much about why the axiom of choice is widely regarded as plausible. Let me try to address that question.

First let's dispose of some non-reasons. In response to your questions, I don't know of anyone who thinks that the Banach–Tarski paradox is a reason to believe in the axiom of choice. I also don't know of anyone who argues, "It is a priori plausible that there exist non-measurable sets; so the fact that the axiom of choice yields the attractive conclusion that there are non-measurable sets is a point in favor of believing the axiom of choice." Instead, those who are comfortable with the existence of non-measurable sets typically start by accepting the axiom of choice, and then they accept non-measurable sets as "part of the territory" that comes with the axiom of choice.

Those who think that Banach–Tarski casts doubt on the axiom of choice typically have a philosophical predisposition that math is supposed to model the physical world closely. So for example, $\mathbb R^3$ is not just a random mathematical structure that we study purely for its own sake; it is supposed to be a decent model of physical space (or at least, open subsets of $\mathbb R^3$ are supposed to model localized regions of physical space). Banach–Tarski, when given a direct physical interpretation in this way, yields something that we "know" makes no physical sense, and so if we think that math is supposed to yield physical truth in this way, then Banach–Tarski is going to lead us to reject something in the math. Whether that "something" we reject is the axiom of choice is a separate question, and Andrej Bauer's excellent answer shows that there are other options, but the point I want to highlight is that we're going to be led down this path in the first place only if we have certain presuppositions about how math and physics are supposed to relate.

There are others who don't view set theory in this way. According to them, set theory is supposed to be about abstract collections of things, and the way to arrive at axioms is by abstractly thinking about what properties they should have, not by comparing them with the physical world. The axiom of choice can be thought of as saying that if you have a bunch of nonempty collections of things, then there is another collection of things that contains one element from each of your original nonempty collections. Stated this way, the principle sounds intuitively plausible, and I would argue that this intuitive plausibility is, at least implicitly, the main argument in the minds of most people who accept the axiom of choice. If this is the way you think, then non-measurable sets and Banach–Tarski are not going to dissuade you from accepting the axiom of choice. Those phenomena will just lead you to say that we can't arrive at physical predictions from mathematics in such a naive manner; instead, to do physics, we have to formulate physical theories. Math can of course help a lot with the construction of physical theories, but it's not as simple as just saying that the mathematical theory of $\mathbb R^3$ is our theory of physical space.

These two options aren't the only options. The work of Solovay shows that you can, to a large extent, have your cake and eat it too, by working in a set-theoretic universe where all subsets of $\mathbb R^n$ are Lebesgue-measurable and a weakened, but still quite strong, version of the axiom of choice known as "dependent choice" is available. Why Solovay's model hasn't become more popular is not completely clear, but perhaps part of the reason is that it feels like a "compromise position," and the people in the two different camps above have not seen any need to migrate to that kind of compromise.

Timothy Chow
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    My answer at http://mathoverflow.net/q/34863 is relevant to the last sentence of your answer here. – Andreas Blass Jan 20 '17 at 23:46
  • The axiom of choice is in contradiction with logic. In order to well-order a set you have to distinguish all its elements. That means you have to give a finite name of its own to each one. That is impossible for an uncountable set. The reason for maintaining the axiom is simply that otherwise great parts of set theory would break down. –  Aug 05 '17 at 15:16
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    @user112009 Following your reasoning, do you have to distinguish and name all (uncountably many) real numbers to put them into their usual linear order? Or do you have to distinguish and name all (uncountably many) countable ordinals to put them into their usual well-order? – Vladimir Reshetnikov Feb 07 '18 at 21:14
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    @user112009 It sounds like you think that the problem with the axiom of choice is that it claims that certain "undefinable" operations are possible. But for example, in $L$, there is a definable well ordering of the universe, and everything you do with the axiom of choice is going to be definable. So this type of criticism of the axiom of choice is unconvincing. Joel David Hamkins explains this point well in another MO answer. – Timothy Chow Dec 01 '22 at 05:48
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Are there reasons why it is plausible (for physicists, philosophers, mathematicians) to believe that not all sets should be measurable?

Yes. If every set of reals is Lebesgue measurable, then you can partition $\mathbb{R}$ into more than continuum many pairwise disjoint non-empty pieces. (See this answer and comments for the details.)

Surely the Banach-Tarski paradox seems unintuitive. But having a set that can be broken up into more pieces than there originally were... is just wrong.

Burak
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    Well of course this will happen if you accept something as barbarous as excluded middle :D – მამუკა ჯიბლაძე Jan 20 '17 at 21:03
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    That's a great (informal) motivating example. I also refuse to accept that the Cartesian product of a family of non-empty sets can turn out empty. Surely it must have at least one element! And Banach–Tarski construction on closer inspection seems to be no more troubling than having a circle with a countable number of points removed from it that becomes a proper subset/superset of itself if we simply rotate it. There is a great visualization of if: https://www.youtube.com/watch?v=s86-Z-CbaHA – Vladimir Reshetnikov Feb 07 '18 at 21:33
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    IMO what this shows is not that "every set of reals is Lebesgue measurable" is counterintuitive, but that the notion of "more than" behaves poorly in the absence of AC. Suppose there exists an injection from $A$ to $B$, and there exists a surjection from $A$ to $B$, but there is no bijection from $A$ to $B$. It's a bit strange to interpret this as saying that $B$ is bigger than $A$. If we lack the tools to construct maps that our intuition about size tells us "should" exist, then IMO we should just admit that our theory of size is inadequate, not that $B$ is really bigger than $A$. – Timothy Chow May 17 '19 at 02:10
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    The issue is that if you assume that every set of reals is Lebesgue measurable, you have decided that your notion of set and "quantity of elements" is geometric/measure theoretic. Cardinality is completely unrelated to geometry, so there is no longer any reason why you should have a single well defined notion of cardinality for uncountable sets. The only thing it tells you is that functions that rely on choice to exist have nothing to do with geometry or measures. – saolof Sep 19 '22 at 06:29
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Arguments from physics may not help. Here is Bryce DeWitt reviewing Stephen Hawking and G.F.R. Ellis using the axiom of choice in 1973:

The book also contains one failure to distinguish between mathematics and physics that is actually serious. This is in the proof of the main theorem of chapter 7, that given a set of Cauchy data on a smooth spacelike hypersurface there exists a unique maximal development therefrom of Einstein’s empty-space equations. The proof, essentially due to Choquet-Bruhat and Geroch, makes use of the axiom of choice, in the guise of Zorn’s lemma. Now mathematicians may use this axiom if they wish, but it has no place in physics. Physicists are already stretching things, from an operational standpoint, in using the axiom of infinity.

It is not a question here of resurrecting an old and out-of-date mathematical controversy. The simple fact is that the axiom of choice never is really needed except when dealing with sets and relations in non-constructive ways. Many remarkable and beautiful theorems can be proved only with its aid. But its irrelevance to physics should be evident from the fact that its denial, as Paul Cohen has shown us, is equally consistent with the other axioms of set theory. And these other axioms suffice for the constructions of the real numbers, Hilbert spaces, C* algebras, and pseudo-Riemannian manifolds–that is, of all the paraphernalia of theoretical physics.

In “proving” the global Cauchy development theorem with the aid of Zorn’s lemma what one is actually doing is assuming that a “choice function” exists for every set of developments extending a given Cauchy development. This, of course, is begging the question. The physicist’s job is not done until he can show, by an explicit algorithm or construction, how one could in principle always select a member from every such set of developments. Failing that he has proved nothing.

Some physicists want to use the axiom of choice, but some physicists don't.

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    It's worthwhile to mention that DeWitt's criticism to the result of Choquet-Bruhat and Geroch no longer applies. Jan Sbierski recently proved the existence of a maximal Cauchy development without using Zorn's lemma: On the Existence of a Maximal Cauchy Development for the Einstein Equations - a Dezornification, in Annales Henri Poincaré 17 (2016) 301-329. – Pedro Lauridsen Ribeiro Jan 20 '17 at 14:21
  • As for the lack of need for the axiom of choice in the "paraphernalia of theoretical physics", as DeWitt puts it, one should keep in mind that this statement may hold true for the definitions of the mathematical objects he lists, but not necessarily for the ensuing results. A typical example is the spectral theorem - a cornerstone of both Hilbert space and C*-algebra theories, and fundamental to quantum mechanics -, which relies on the Stone-Weierstrass theorem (for the continuous functional calculus) and also on the Riesz representation theorem (for the $L^\infty$ functional calculus). – Pedro Lauridsen Ribeiro Jan 20 '17 at 14:46
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    It's interesting to note, though, that by loosening the notion of "space", as @AndrejBauer puts it in his answer, one can indeed make the spectral theorem for commutative C-algebras (a.k.a. Gelfand duality) constructive. More precisely, one replaces "compact Hausdorff topological space" by "compact, completely regular locale in a Grothendieck topos" in the statement of Gelfand duality - this was shown by B. Banaschewski and C.J. Mulvey in A Globalisation of the Gelfand Duality Theorem, Ann. Pure Appl. Logic 137* (2006) 62-103. – Pedro Lauridsen Ribeiro Jan 20 '17 at 14:54
  • @PedroLauridsenRibeiro Yes, many theorem-proving physicists like the spectral theorem; other physicists do not consider any theorems fundamental; de Witt would probably be in the middle. Saying that this theorem is fundamental to quantum mechanics is making a judgment about physics and the use of math in it, of the same sort as our judgments about math and the use of choice in it. –  Jan 20 '17 at 16:05
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    Some physicists want to use the axiom of choice, but some physicists don't. Based on Pedro Lauridsen Ribeiro's comment, I doubt that this is the case. I doubt that any physicists, think the axiom of choice is relevant to physics, but there may be cases where it's a nice convenience, and they may not know how to or want to bother with eliminating choice from their arguments. –  Jan 20 '17 at 17:53
  • What would such a critic make of the law of the excluded middle? I don't know physics, but “an explicit algorithm or construction” wouldn't use LEM in my book. – mudri Jan 20 '17 at 19:00
  • @BenCrowell, given what de Witt describes, would you count Hawking as a physicist who thinks the axiom of choice is relevant to physics? –  Jan 20 '17 at 19:11
  • @JamesWood, I agree with you that de Witt's argument could go against $LEM$. Taking it further: given a theorem $T$, do physicists ever need to prove $T$? Could they do physics equally well using only $ZFC \vdash T$, as a theorem of $PRA$ which does not require $LEM$ or believing anything about sets? –  Jan 20 '17 at 19:20
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    @Pedro: Thanks for pointing out the new AHP article. However, I disagree about the spectral theorem (even for finite collections of commuting unbounded self-adjoint operators) needing the AC. As long as the Hilbert space is separable, it can be done with ACDC only. See this discussion on physics.SE http://physics.stackexchange.com/questions/43853/does-the-axiom-of-choice-appear-to-be-true-in-the-context-of-physics/44057#44057 – Abdelmalek Abdesselam Jan 20 '17 at 20:20
  • @MattF.: given what de Witt describes, would you count Hawking as a physicist who thinks the axiom of choice is relevant to physics? The fact that Hawking and Ellis used the axiom of choice doesn't tell us anything about their philosophical stance, if any. They probably used it simply as a convenience. Pedro Lauridsen Ribeiro's comment tells us that the proof holds without the axiom of choice, but it sounds like it's just a lot more work that way. –  Jan 20 '17 at 20:27
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    @JamesWood: Physicists have been using infinitesimals since Newton. If you take any argument that uses infinitesimals, you can translate it into the language of either non-standard analysis or smooth infinitesimal analysis. NSA uses aristotelian logic, while SIA uses nonaristotelian logic. I think this is a good example of how utterly irrelevant such foundational issues are in physics. Physicists have a body of techniques that have been shown to be valid by comparison with experiment, not by showing their (self-)consistency within some foundational framework that is in style. –  Jan 20 '17 at 20:38
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    Whenever constructibility is brought up, I always feel compelled to point out that the axiom of choice is the theorem of choice if you limit yourself to such. More precisely, $ZF+(V=L)\vdash C$ –  Jan 20 '17 at 22:48
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    @AbdelmalekAbdesselam Agreed, also because Gelfand duality is actually equivalent to the Boolean prime ideal theorem, which is a bit weaker than AC. Nonetheless, the axiom of (countable) dependent choices (normally acronymed DC) is still non-constructive. A large part of analysis can indeed be done just with DC (Baire's theorem, for instance, is equivalent to it), but DeWitt's criticism still stands - he begs for an actual construction of the choice function, even if it is a countably dependent one. Thanks for the physics.SE link, I'll have a look. – Pedro Lauridsen Ribeiro Jan 21 '17 at 01:02
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    This also begs the question of whether separable Hilbert spaces suffice for applications, specially regarding (mathematical) physics: physics.stackexchange.com/questions/90004/separability-axiom-really-necessary/ – Pedro Lauridsen Ribeiro Jan 21 '17 at 01:02
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    One can get away with a strict form of finitism as one's mathematical foundation for physics, as shown by Ye (https://dx.doi.org/10.1007/978-94-007-1347-5, free draft copy here: http://www.phil.pku.edu.cn/cllc/people/fengye/finitismAndTheLogicOfMathematicalApplications.pdf). From a review [1]: "most of the mathematics necessary for modern theoretical physics can be developed within a strict finitist framework. In Chapter 8, Ye outlines semi-Riemannian geometry sufficient for proving a version of Hawking’s singularity theorem." ([1] http://journals.uvic.ca/index.php/pir/article/view/13181/4184) – David Roberts Jan 21 '17 at 01:06
  • @DavidRoberts, if the physicists say "I don't like Ye's interpretation of my math" or "I don't need a mathematical foundation for my physics at all", how does an appeal to physics settle any of the mathematical questions at issue? Or more briefly: "necessary" by what standard? –  Jan 22 '17 at 03:16
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    @MattF. I would almost claim that one can't appeal to physics to justify infinitary mathematics, because physics doesn't need that logical strength. Recall that mathematics for years didn't itself use a foundation, and barring some corner cases, worked perfectly fine. – David Roberts Jan 22 '17 at 06:42
  • @DavidRoberts, I agree with that comment, but it seems to mean your earlier comment, "one can get away with finitism", has little force. –  Jan 22 '17 at 16:52
  • @MattF. Sorry, 'get away with' is a very imprecise way of saying what I said in the second comment, second half of the first sentence. – David Roberts Jan 23 '17 at 10:07
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    @Pedro: I find ACDC (I like this acronym) more intuitive than the full-fledged AC and it agrees with my standards of constructive math. If one gives up on it then there is not much analysis that survives, in particular Baire's Thm as you pointed out. I think it si OK to use AC for the first proof of some result like the on in general relativity you mentioned but then this should be considered a temporary fix awaiting a more satisfying constructive proof (eventually using ACDC). – Abdelmalek Abdesselam Jan 23 '17 at 15:23
  • @AbdelmalekAbdesselam "I find ACDC (I like this acronym)..." - I see what you did there... Anyway, concerning the (mathematical) physicist's stance on constructibility in mathematics, I had a similar discussion with an OP in the comments to my MO answer http://mathoverflow.net/questions/238153/physical-meaning-of-the-lebesgue-measure/238174#238174 – Pedro Lauridsen Ribeiro Jan 23 '17 at 16:42
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Just a quick addendum: the result was actually found by Felix Hausdorff and then repackaged in a more spectacular form by Banach and Tarski. Hausdorff' point was precisely to show that the axiom of choice leads to such unreasonable consequences that it should probably be avoided. This was a hot topic of discussion during the following decades, and it seems that the answer at large from the mathematical community is: ok, there are inconveniences, but the advantages of using the AC are superior and we prefer to have it available.

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    Did Hausdorff really advocate avoiding the axiom of choice? He used the axiom of choice in some of his important work, for example the construction of Hausdorff gaps. – Andreas Blass Jan 20 '17 at 23:51
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    If I recall correctly, he was at the time skeptical, but after just a few years he was converted and became a supporter of the use of the AC – Piero D'Ancona Jan 21 '17 at 00:18
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I think one of the strongest arguments for the axiom of choice is that every model of ZF contains as an inner model a constructible universe $L$, and $AC$ is a theorem of the constructible universe. We have

$$ ZF+(V=L) \vdash AC $$

In other words, a necessary condition for asserting $\neg AC$ is to first assert "there exists a set that cannot be constructed" — that is, it requires positing the existence of additional structure above and beyond what is guaranteed by ZF.

Thus, it seems clear to me that ZFC is by far the better choice for foundations. One may still wish to work with another set theoretic universe, but that's most appropriately done as the study of additional structure built atop of the foundations, not by rewriting the foundations themselves.

  • @Burak: Same reason: every model contains an $L$. Assuming $V \neq L$ is assuming extra structure built atop of $L$, so taking $L$ is better for foundations. –  Jan 20 '17 at 23:21
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    Every model of ZF contains a model of ZF - Infinity. Is that a good argument for finitism? – David Roberts Jan 21 '17 at 00:39
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    I agree with David Roberts. This is an unconvincing argument for AC. The argument amounts to this: If we restrict ourselves to a very limited notion of set, namely constructible sets, then there is always a choice function for a constructible family of constructible sets, and in fact the choice function is constructible. Therefore (?!) choice functions surely (?!) exist for arbitrary families of sets. – Timothy Chow Jan 21 '17 at 03:16
  • @Timothy: That the universe of sets we take for foundations satisfies AC by no means implies the other universes we may wish to study must also satisfy AC. –  Jan 21 '17 at 04:57
  • @David: Can I study universes of sets satisfying, say, ZFC, if I take ZF-I as foundations? My impression is no. Am I wrong? Taking ZFC+(V=L) as a starting point, on the other hand, is still rich enough to allow construction of lots of other interesting universes. –  Jan 21 '17 at 05:00
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    @Hurkyl: I think you misunderstand my point. That $ZF + (V=L) \vdash AC$ tells us only that $AC$ holds if we restrict ourselves to constructible sets. That (to me) does not suggest that $AC$ holds for all sets, which is what $ZFC$ says. If we accept your line of argument, then it would also be an argument for the generalized continuum hypothesis, and are you really going to argue that? This only seems convincing to me if you're going to argue that $V=L$ is true. – Timothy Chow Jan 21 '17 at 05:28
  • @Timothy; ZFC only says that AC holds in models of ZFC. –  Jan 21 '17 at 05:30
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    @Hurkyl : No, ZFC says that AC holds, period. That AC holds in models of ZFC is practically a tautology. – Timothy Chow Jan 21 '17 at 05:33
  • @Timothy: Okay. In ZFC, I construct the arrow category $X \to Y$. Since that arrow is an epimorphism, "AC holds, period" implies it is split. But I don't see any arrows from $Y$ to $X$ in the arrow category.... Having AC in the ambient set theory does not imply that other structures you construct satisfy their internal versions of AC. –  Jan 21 '17 at 05:40
  • Who knows? I think Timothy Chow is making some valid points, however. I'm generally agnostic regarding foundations (the best mathematics is generally foundation-independent, give or take) but Maddy's arguments about V=L being an impoverishment of the universe of sets are quite striking. – David Roberts Jan 21 '17 at 06:08
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    @Hurkyl: I still don't understand why this is supposed to be an argument for $AC$. Do you think the following is an argument for $AC$? "We have $ZF+GCH \vdash AC$ and hence in order to assert $\neg AC$ one should posit the existence of additional structure on cardinal numbers other than expected. Also, $ZF+GCH$ as a starting point is rich enough to allow constructions of other interesting universes. etc." Your whole argument can be recast with $V=L$ replaced by $GCH$. – Burak Jan 21 '17 at 09:35
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    @Burak : Although I'm not convinced by Hurkyl's argument, I don't think that you can replace V=L by GCH. The point is that L is essentially the minimal model for ZF. Hurkyl's point is that AC is true if you take the smallest possible universe of sets that are forced on you by ZF. – Timothy Chow Jan 22 '17 at 21:31
  • You say $ZF + (V=L) \vdash AC$. But that claim of probability can be proved in PRA. Why then do you say that the better choice for foundations is something like ZFC rather than something like PRA? –  Jan 22 '17 at 22:10
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    @MattF. : Again, I'm not defending Hurkyl's argument, but the thesis isn't that $ZFC$ is better than $PRA$; rather, the thesis is that $ZFC$ is better than $ZF + \neg AC$, because $ZF + \neg AC$ implies something that seems unjustified (namely, that $V\ne L$). The fact that this argument cites a fact that happen to be provable in $PRA$ doesn't at all suggest that we should use $PRA$ for foundations; that's a non sequitur. – Timothy Chow Jan 23 '17 at 16:17
  • @TimothyChow, the question is what concept of foundations is used here. (Note that the higher-voted answers do not rely on that word.) Until I see that articulated a bit more, it's Hurkyl's argument that looks like a non-sequitur to me. –  Jan 23 '17 at 20:24
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    @MattF. : Regardless of what "foundations" means, it is clear from context that when Hurkyl is saying that $ZFC$ is a "better choice", the intended assertion is that $ZFC$ is a better choice than $ZF + \neg AC$. Dragging $PRA$ into the picture out of left field is the non sequitur. – Timothy Chow Jan 25 '17 at 15:44
  • Is not it evident that there are sets that cannot be constructed? – Anixx Jun 18 '21 at 23:20
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Physical applications of the Banach–Tarski theorem were explored by Henry Kuttner, The Time Axis, Startling Stories 18:3 (January 1949), 13–82. Some excerpts from pp. 66–67:

"'Professor Raphael M. Robinson of the University of California now shows that it is possible to divide a solid sphere into a minimum of five pieces and reassemble them to form two spheres of the same size as the original one. Two of the pieces are used to form one of the new spheres and three to form the other.

"'Some of the pieces must necessarily be of such complicated structure that it is impossible to assign volume to them. Otherwise the sum of the volumes of the five pieces would have to be equal both to the volume of the original sphere and to the sum of the volumes of the two new spheres, which is twice as great.'"

[. . . .]

"This is it," he said.

Even the crowd around the neural-web table thinned as the workers in the laboratory flocked around him to watch.

He had a sphere about the size of a grapefruit, floating in mid-air above his table. He did things to it with quick flashes of light that acted exactly like knives, in that it fell apart wherever the lights touched, but I got the impression that those divisions were much less simple than knife-cuts would be. The light shivered as it slashed and the cuts must have been very complex, dividing molecules with a selective precision beyond my powers of comprehension.

The sphere floated apart. It changed shape under the lights. I am pretty sure it changed shape in four dimensions, because after a while I literally could not watch any more. The shape did agonizing things to my eyes when I tried to focus on it.

When I heard a long sigh go up simultaneously from the watchers I risked a look again.

There were two spheres floating where one had floated before.

bof
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I think it is interesting to re-phrase this question relative to other axioms and/or theorems of ZFC. What Andrej Bauer's answer suggests is that it may not be the axiom of choice per se that is the culprit, but rather the underlying structure.

For example, it is provable that the existence of non-Lebesgue measurable sets and the Banach–Tarski paradox are both an implication over ZF (without Choice) of the Hahn–Banach theorem (HB) in functional analysis. This means that we can prove those "pathologies" as a theorems from ZF+HB.

Another way to put it is to analyze the structure within the context of different orders of logic. In ordinary first-order logic Choice is provably equivalent to the well-ordering theorem (WO) over ZF. What is different in second-order logic is that WO is strictly stronger than choice: WO $\vdash_{ZF}$ Choice, but Choice $\nvdash_{ZF}$ WO.

In other words, one may get the feeling that perhaps there has been too much emphasis put on Choice. Other axioms and theorems can prove just as problematic. The approach described above is the essence of Reverse Mathematics (RM). While RM has been traditionally carried out at a much lower proof-theoretic strength level (subsystems of second-order arithmetic), in my opinion this provides a very useful framework for analyzing the foundations of other parts of mathematics.

Dawid K
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    Your statement that the existence of non-Lebesgue measurable sets is equivalent over ZF to the Hahn-Banach theorem is surprising to me. Could you give a reference (or a proof) for the implication from non-measurable sets to HB? – Andreas Blass Jan 27 '17 at 19:05
  • @AndreasBlass To my knowledge the original result is due to Foreman and Wehrung in the following paper from Fundamenta Mathematicae (1991): http://matwbn.icm.edu.pl/ksiazki/fm/fm138/fm13812.pdf – Dawid K Jan 27 '17 at 23:12
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    Ah, I see what you did there. The word equivalent is misleading, since the proved implication is only from HB to non-measurable sets, and not the other way. Amending the original answer. – Dawid K Jan 27 '17 at 23:25
  • It may also be worth mentioning that HB is related to the ultrafilter lemma rather than "full" choice. But of course the ultrafilter lemma is enough to prove non-measurable sets and is arguably the part of choice most important for that. IIRC, the BPIT equivalent to Tychonoffs theorem for hausdorff spaces which is sufficient to prove HB. – saolof Sep 20 '22 at 07:30