I have only recently learned set theory as a side hobby (I'm a mathematical physics undergraduate), and I have heard various Profs and friends that are math graduates just ridicule or dismiss the idea of rejecting the axiom of choice. I don't know enough to formulate a good argument for rejecting the axiom of choice (or don't even know if a good argument for rejecting it exists). What is a good one for rejecting it? And if you would like to rebuttal it too that would be helpful.

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    $\begingroup$ See The Axiom of Choice for a first overview. $\endgroup$ Jun 13 '19 at 14:49
  • $\begingroup$ It seems pretty intuitive that if we have a collection of non-empty disjoint sets, we should be able to take an element from each and form a new set. $\endgroup$
    – John Douma
    Jun 13 '19 at 14:50
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    $\begingroup$ There is the Banack-Tarski paradox, which could be considered a reductio ad absurdum of the Axiom of Choice. $\endgroup$ Jun 13 '19 at 14:56
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    $\begingroup$ There are no good arguments. The Banach Tarski is not a good argument for abandoning choice. $\endgroup$
    – Asaf Karagila
    Jun 13 '19 at 14:56
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    $\begingroup$ What does it mean to "reject" the axiom of choice? If I reject AC, does that mean that I consider the consequences of the theory ZF + AC not worth studying? Or does it just mean that I am interested in studying the theory ZF + not-AC? $\endgroup$
    – bof
    Jun 13 '19 at 16:15

One good argument could be that if you want to ensure that every set of reals is Borel. This can happen if the real numbers are a countable union of countable sets. So that means that you're too tired of those proofs that a countable union of countable sets is countable.

Okay, so you're willing to take Dependent Choice and countable choice on the chin. That's fine. Then not all sets of reals are Borel. But a good argument would be that you want all sets of reals to be Lebesgue measurable. That's it. No more pesky Banach–Tarski. You cannot partition the unit ball in weird ways. Except that you can partition the unit ball into strictly more non-empty sets than points.

I mean, really, there's no good argument for rejecting the axiom of choice. There are several ad hoc arguments that might be of interest.

  1. All sets of reals are Lebesgue measurable, or at least with the Baire Property. In that scenario all linear operators between a completely metrizable group and a normed group are continuous. This means that all linear operators between Banach spaces are continuous, which admittedly can be pretty nice.

    Cost: No Hahn–Banach, no Banach–Alaoglu, no ultrafilters, no compactness/completeness theorem for first-order logic.

  2. If you want to go deeper into the rabbit hole as far as set theoretic ideas go, you need to reject the Axiom of Choice in order to take into your heart the Axiom of Determinacy. Even then, though, most people would argue that we really study inner models where it holds and $\sf AC$ holds in the full universe.

    You can, however, decide to study very large cardinals. Reinhardt and Berkeley cardinals, and their relatives. With those you really have to reject $\sf AC$ pretty much wholesale. But do they have consequences "downstairs" at the working mathematician's level of the set theoretic universe? Not as much. Not directly. Not things we cannot prove with far weaker (and choice-compatible) axioms instead.

All reasons to reject choice are really reasons to reject infinite sets and the law of excluded middle. And they are not good reasons for that either.

  • $\begingroup$ "Except that you can partition the unit ball into strictly more non-empty sets than points" Should this be called Banach-Tarski's revenge? $\endgroup$
    – user76284
    Jun 15 '19 at 19:35
  • $\begingroup$ It's called the division paradox. $\endgroup$
    – Asaf Karagila
    Jun 15 '19 at 20:06

Earnest opposition to the axiom of choice is mostly a thing of the early 20th century. At that time the Zermelo-Fraenkel axioms for set theory were pretty new, and the circumstantial evidence that they succeed in avoiding the set-theoretic paradoxes was not as overwhelming as it is today. In that situation, approaching proposed additional axioms with a sound caution was understandable.

The argument from caution goes mostly like this: On one hand, the axiom of choice has some rather surprising and non-intuitive consequences, such as Banach-Tarski. If something like that -- which certainly doesn't correspond to any practical truth about physical space, which we'd like for $\mathbb R^3$ to model -- can be proves, who's to say that it's not because it proves everything, including contradictions? If it did, that would be bad indeed.

On the other, hand the philosophical motivation for the axiom of choice is somewhat weak. Sure enough it corresponds to a form of reasoning that had long been considered unproblematic until the proper foundation of set theory became urgent. But on the other hand, all the other axioms corresponded (at least morally) to operations where it seems to be pretty clear that there's one and only one correct result. On the the axiom of choice purports to give you something that is -- by design -- extremely under-specified. So in an argument that involves it, it is no longer clear how you would distinguish a correct result from a wrong one.

(The latter part of this argument partly ignores that usual first-order logic does allow us to make single choices along the way in a proof, and those choices can be pretty underspecified too. But making an infinity of them seems to have greater psychological impact, especially if you're already skeptical of "completed infinities" and suspect that the root of the paradoxes was basically related to playing too hard and fast with infinity).

However, this whole argument from caution came tumbling down when Gödel proved (using techniques that don't depend on AC itself, or even on infinitary reasoning at all) that if ZF is consistent then ZFC is consistent too. In other words, assuming AC will not lead to any increased chance of inconsistency.


If the other usual set-theory axioms (which are called ZF) are consistent, they can neither prove nor disprove AC, so we have to decide whether we believe it intuitively. AC implies some strange results concerning non-measurable sets, such as the possibility of rearranging a sphere into two of the same size. This is related to the fact that, by inspiring some impressive things exist, AC implies you can do some very impressive things.

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    $\begingroup$ I find your use of intuitive odd. I think a lot of people would argue that AC is in fact intuitively true, since it is intuitively clear how to produce a choice function... Choose an element from each set! $\endgroup$
    – Asaf Karagila
    Jun 13 '19 at 17:23
  • $\begingroup$ @AsafKaragila I'm an AC supporter attempting to sate the OP by playing devil's advocate. If I had to summarise the role of intuitions on both sides in one sentences, I'd write: it's intuitive that choice functions exist, but the consequences of this run counter to at least the intuitions of AC detractors. That's the trouble with intuitions; even those of one person can be incongruous. $\endgroup$
    – J.G.
    Jun 13 '19 at 17:26

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