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If I understand correctly, AC is not needed in the case of a finite union of countable sets because the choice function can be built with recursion, using the case of 2 sets (which is straight-forward). However, why is AC needed in the case of $\omega$ countable sets (or $\alpha$ infinite (well-ordered) sets, for $\omega\leq\alpha\in\text{Ord}$)?

If we take the choice function $G$ to be $\bigcup_{n<\omega}G_n$, where $G_n$ is the choice function for the case of $n$-sets union, such that $\forall m<n \;\; G_m\subseteq G_n$, then $G$ will also be a function and a choice function.

Where's my mistake/where do I implicitly use AC?

Thanks

Idan
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  • You are choosing each $G_n$ among many possible choice functions, which is infinitely many choices – Alessandro Codenotti Jun 28 '20 at 12:10
  • $G_n$ is the choice function for a union of $n$ countable sets, and it can be chosen (with recursion) like in the case of a union of 2 countable sets, no? Given $G_{n-1}$, $G_n$ will be $G_{n-1}$ union a bijective function $A_n\to\mathbb N$ (where $A_n$ is the n'th set), which can be chosen by definition – Idan Jun 28 '20 at 12:15
  • Also, https://math.stackexchange.com/questions/717961/countable-axiom-of-choice-why-you-cant-prove-it-from-just-zf https://math.stackexchange.com/questions/2816939/when-induction-proofs-need-a-choice-function https://math.stackexchange.com/questions/307277/why-isnt-this-a-valid-argument-to-the-proof-of-the-axiom-of-countable-choice and many many others dealing with these sort of arguments. – Asaf Karagila Jun 28 '20 at 12:20
  • @idan What you can do with arbitrarily large finite sets isn't necessarily possible with infinite ones. – J.G. Jun 28 '20 at 12:28
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    Also to clarify on your second comment, you're saying "$G_n$ the choice function ...", but there is no the choice function. There are many choice functions. – Asaf Karagila Jun 28 '20 at 14:03
  • My mistake. However I read the other links and what I understood is: if it's finite it's possible, if not you need the AC "just because" (but there's got to be a more logical reason). If it's needed because arbitrary choice is not trivial, then why do finite union (and binary union) of countable sets doesn't need AC? And if the latter is true, then what fails when taking the union of $G_n$? They are functions (monotonic by $\subseteq$) that I constructed using recursion on ordinals, so the union should also be a function – Idan Jun 28 '20 at 14:27
  • Because you can make finitely many choices. In addition to all the links so far, see also https://math.stackexchange.com/questions/1839913/axiom-of-choice-where-does-my-argument-for-proving-the-axiom-of-choice-fail-he and https://math.stackexchange.com/questions/365269/intuition-behind-the-axiom-of-choice – Asaf Karagila Jun 28 '20 at 17:59

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