# Axiom of Choice for finite vs infinite product of sets

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 $$A$$ ?

• 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 – miraunpajaro Aug 29 '19 at 13:35
• 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? – Asaf Karagila Aug 29 '19 at 14:47
• 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). – Andreas Blass Aug 29 '19 at 18:42

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.