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Consider the equivalence relation on $[0,1]$ where $$x \sim y \iff x - y \in \mathbb{Q}.$$ Fix a set $S$ which contains exactly one representative from each equivalence class of the relation. It is well-known that constructing such a set requires the axiom of choice. However, I don't have an intuition for what breaks down if the axiom of choice is not assumed.

If I don't assume the axiom of choice, what happens to $S$? Does it cease to exist? If not, then surely $S$ is some subset of $[0,1]$, whether or not we invoke the axiom of choice, so why can't I refer to $S$ in a proof without choice, in the same way that I would refer to any other subset of $S$?

I haven't studied set theory formally, so please excuse my ignorance.

TheProofIsTrivium
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    In the Solovay model of set theory every set of reals is Lebesgue measurable, so no such $S$ exists. – Brian M. Scott Sep 26 '20 at 06:25
  • My understanding is that such a set would still exist, but we wouldn't be able to describe/refer to it. The axiom of choice allows us to work with such sets. However, I will let someone who understands this better post an answer – JasonM Sep 26 '20 at 06:25
  • @BrianM.Scott Why wouldn't this set be Lebesgue measurable? – JasonM Sep 26 '20 at 06:27
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    @JasonM: It’s a Vitali set, the standard example of a non-measurable set of reals. – Brian M. Scott Sep 26 '20 at 06:28
  • @BrianM.Scott I'm sorry, I'm confused. If the set doesn't exist, how is it an example of a non-measurable set? – JasonM Sep 26 '20 at 06:32
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    @JasonM: It does exist if one assumes the axiom of choice, which one pretty routinely does unless one is specifically interested in what can be done without it. – Brian M. Scott Sep 26 '20 at 06:35
  • @JasonM: Do you understand why it doesn't exist without axiom of choice, or not ? – Bruce Sep 26 '20 at 06:38
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    @BrianM.Scott Okay, I think I just misunderstood what you said. Apologies – JasonM Sep 26 '20 at 06:42
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    @Bruce: It doesn’t actually need the full strength of $\sf{AC}$: there are models in which $\sf{AC}$ is false but there is a Vitali set. – Brian M. Scott Sep 26 '20 at 06:46
  • According to Asaf Karagila in this question a class that is an element of a set is a set, so $S$ does exist. It's the function that maps from the equivalence classes to representatives that requires $\mathsf{AC}$. – Chrystomath Sep 26 '20 at 06:54
  • For most questions, I change the tag set theory to elementary set theory, but this time, I added the tag set theory, as I feel it is relevant. – J.-E. Pin Sep 26 '20 at 07:58
  • The question of whether or not S exists does not make sense by itself. The right question is "Does S exist in a particular model of set theory?" If the model satisfies AC then S exists in that model. If the model is the Solovay model then S does not exist in that model. Moreover, there are models where S exists and AC fails (see: https://math.stackexchange.com/questions/861134/does-vitali-set-imply-the-axiom-of-choice). – halrankard2 Sep 26 '20 at 14:26

1 Answers1

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Nothing "happens" to $S$; but you might simply not be able to prove such an $S$ exists.

Note your description of $S$ : it's a set that contains one representative of each equivalence class. Why should it exist ?

Saying "Fix such a set $S$" requires a proof that it exists.

Let me start with an easier example: suppose you're working with some set of numbers, but you don't know whether it's $\mathbb R$ or $\mathbb C$. You're also given some $x$, and you want to say "fix a $y$ be such that $y^2=x$" : why does it exist ? If you know you're working in $\mathbb C$ (analogous to assuming the axiom of choice), then there's a theorem (analogous to the axiom of choice) which tells you that such a $y$ exists, so you're well and good.

But if you don't know and might actually be working in $\mathbb R$ (analogous to being in a situation where AC fails), you can't be sure that such a $y$ exists, and so you can't prove it; and so you can't "use" the $y$ (of course you can still prove properties about such $y$'s, just as here you can prove things on such $S$'s, but that tells you nothing about whether they exist !)

Back to your situation: one point of view on the statement that AC cannot be proved from ZF (assuming ZF is consistent) is that there are "universes of math" where AC simply does not hold. In such universes, it would be possible that no such $S$ exists. It didn't "cease" to exist, it simply never did. But there are other "universes of math" where such an $S$ might exist.

If you could give an explicit definition of $S$, then you could always refer to it (although, it might be non obvious why this definition actually defines a set).

Maxime Ramzi
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