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Why we don't need Axiom of Choice to prove the following statement

Let $S_{\alpha}, \alpha \in A$ be a family of disjoint nonempty sets, and consider $P = \bigcup_{\alpha \in A} S_{\alpha}$. If $|A|$ is finite then there exists $Q \subset P$ such that for each $\alpha \in A$, we have $|Q \cap S_{\alpha}| = 1$

with this as the proof (taken from https://math.stackexchange.com/a/29383/)

Since each of the $S_\alpha$'s are nonempty, then by definition for each $\alpha$ there exits $b_{\alpha} \in S_{\alpha}$. So $Q = \{b_{\alpha} | \alpha \in A \}$ works.

But apparently we do need Axiom of Choice to prove the exact same hypothesis with just the hypothesis ``$|A|$ is finite" removed.

Can someone provide some intuition on why the proof won't work for infinite $A$ ?

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  • $\begingroup$ In the proof you adjointed there is a step where they use a choice function. So that's why it doesn't work for finite sets $\endgroup$ – miraunpajaro Aug 29 '19 at 13:35
  • $\begingroup$ math.stackexchange.com/questions/85153/… math.stackexchange.com/questions/717961/… and I feel like we've covered this before a lot on the site. Did you go through some of the old questions about this subject? $\endgroup$ – Asaf Karagila Aug 29 '19 at 14:47
  • $\begingroup$ The proof you said was "taken from math.stackexchange.com/a/29383" doesn't seem to actually be there. Furthermore, your formulation of the proof is confusing because it never explicitly uses the finiteness of $A$ (whereas the linked answer to the earlier question does). $\endgroup$ – Andreas Blass Aug 29 '19 at 18:42
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You're choosing the $b_\alpha$. If you can specify a way to single out a unique $b_\alpha$ for each $\alpha$ (including if $|A|$ is finite, and you, say, list the $b_\alpha$ explicitly), the axiom schema of replacement can tell you that $\{b_\alpha\mid \alpha\in A\}$ is a set. If not, then none of the ZF axioms let you say that $\{b_\alpha\mid \alpha\in A\}$ (whatever it now is) is a set. That's exactly what the Axiom of Choice does.

So without Chioce, and with infinite $|A|$, in general a collection of the form $\{b_\alpha\mid \alpha\in A\}$ will not be a set. With Choice, you know that there is at least one such set, although it doesn't say in any way how the $b_\alpha$ in such a set were chosen.

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