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?

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    $\begingroup$ 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. $\endgroup$
    – user78249
    Jan 20, 2017 at 2:16
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    $\begingroup$ 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. $\endgroup$ Jan 20, 2017 at 3:02
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    $\begingroup$ @WillieWong By the way, the idea behind both is due to Hausdorff (essentially, Banach and Tarski only popularized it.) $\endgroup$ Jan 20, 2017 at 3:26
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    $\begingroup$ 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. $\endgroup$ Jan 20, 2017 at 5:26
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    $\begingroup$ 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?" $\endgroup$
    – Drunix
    Jan 20, 2017 at 14:29

9 Answers 9


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.

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    $\begingroup$ 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. $\endgroup$
    – tomasz
    Jan 20, 2017 at 13:25
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    $\begingroup$ @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. $\endgroup$ Jan 20, 2017 at 19:18
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    $\begingroup$ 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? $\endgroup$ Jan 20, 2017 at 20:49
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    $\begingroup$ 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. $\endgroup$ Jan 30, 2017 at 10:23
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    $\begingroup$ 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. $\endgroup$ Jan 30, 2017 at 10:26

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.

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    $\begingroup$ 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. $\endgroup$ Jan 20, 2017 at 14:58
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    $\begingroup$ 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). $\endgroup$
    – Nik Weaver
    Jan 20, 2017 at 15:42
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    $\begingroup$ 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 $\neg x \in S$? $\endgroup$
    – user21820
    Jan 20, 2017 at 15:44
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    $\begingroup$ "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! $\endgroup$
    – Qfwfq
    Jan 20, 2017 at 20:34
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    $\begingroup$ @user21820 what are the elements of that set? What if I eat half of a rook? :-) $\endgroup$
    – David Roberts
    Jan 21, 2017 at 0:43

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.

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    $\begingroup$ My answer at mathoverflow.net/q/34863 is relevant to the last sentence of your answer here. $\endgroup$ Jan 20, 2017 at 23:46
  • $\begingroup$ 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. $\endgroup$
    – user112009
    Aug 5, 2017 at 15:16
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    $\begingroup$ @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? $\endgroup$ Feb 7, 2018 at 21:14
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    $\begingroup$ @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. $\endgroup$ Dec 1, 2022 at 5:48

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.

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    $\begingroup$ Well of course this will happen if you accept something as barbarous as excluded middle :D $\endgroup$ Jan 20, 2017 at 21:03
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    $\begingroup$ 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: youtube.com/watch?v=s86-Z-CbaHA $\endgroup$ Feb 7, 2018 at 21:33
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    $\begingroup$ 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 $A$. $\endgroup$ May 17, 2019 at 2:10
  • $\begingroup$ 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. $\endgroup$
    – saolof
    Sep 19, 2022 at 6:29

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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    $\begingroup$ 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. $\endgroup$ Jan 20, 2017 at 14:21
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    $\begingroup$ 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. $\endgroup$ Jan 20, 2017 at 14:54
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    $\begingroup$ @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. $\endgroup$ Jan 21, 2017 at 1:02
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    $\begingroup$ 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/ $\endgroup$ Jan 21, 2017 at 1:02
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    $\begingroup$ One can get away with a strict form of finitism as one's mathematical foundation for physics, as shown by Ye (dx.doi.org/10.1007/978-94-007-1347-5, free draft copy here: phil.pku.edu.cn/cllc/people/fengye/…). 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] journals.uvic.ca/index.php/pir/article/view/13181/4184) $\endgroup$
    – David Roberts
    Jan 21, 2017 at 1:06

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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    $\begingroup$ 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. $\endgroup$ Jan 20, 2017 at 23:51
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    $\begingroup$ 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 $\endgroup$ Jan 21, 2017 at 0:18

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.

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    $\begingroup$ Every model of ZF contains a model of ZF - Infinity. Is that a good argument for finitism? $\endgroup$
    – David Roberts
    Jan 21, 2017 at 0:39
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    $\begingroup$ 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. $\endgroup$ Jan 21, 2017 at 3:16
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    $\begingroup$ @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. $\endgroup$ Jan 21, 2017 at 5:28
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    $\begingroup$ @Hurkyl : No, ZFC says that AC holds, period. That AC holds in models of ZFC is practically a tautology. $\endgroup$ Jan 21, 2017 at 5:33
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    $\begingroup$ @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 $GCH$. $\endgroup$
    – Burak
    Jan 21, 2017 at 9:35

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.

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    $\begingroup$ This is amusing, but it isn't an answer. $\endgroup$
    – user21349
    Jan 27, 2017 at 17:55

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.

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    $\begingroup$ 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? $\endgroup$ Jan 27, 2017 at 19:05
  • $\begingroup$ @AndreasBlass To my knowledge the original result is due to Foreman and Wehrung in the following paper from Fundamenta Mathematicae (1991): matwbn.icm.edu.pl/ksiazki/fm/fm138/fm13812.pdf $\endgroup$
    – Dawid K
    Jan 27, 2017 at 23:12
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    $\begingroup$ 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. $\endgroup$
    – Dawid K
    Jan 27, 2017 at 23:25
  • $\begingroup$ 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. $\endgroup$
    – saolof
    Sep 20, 2022 at 7:30

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