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It's written in many resources that if we consider only finite sets, that choice axiom can be skipped and be proven from other ZF axioms.

I remember I've read the following explanation. If $A$ is a finite set, there is exists some natural $n, |A|=n$, therefore exists some bijection $f:\{1,\cdots,n\}\to A$. And we can choose an element from $A$ by choosing $f(1)$

But there is a problem with this explanation.

Actually there are $n!$ of such bijections. So how can we choose one $f$ of them without this axiom of choice?!

UPD: actually I can simplify my question heavily.

Let's consider one-point set $A, |A|=1$. How can I get an element from this set without using axiom of choice? We know that $A=\{x\}$ for some $x$. But we don't know how to obtain it. What allows us to just say Let's pick $x$ from $A$?

mnaoumov
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  • If a set of certain objects exists then certainly one such object exists, yes? – basket Jul 01 '16 at 09:05
  • @basket, yes it certainly exists, but I think you cannot use it further without providing an algorithm of obtaining it. From my understanding that's what axiom of choice's role is – mnaoumov Jul 01 '16 at 09:13
  • The existential claim is entirely the Axiom of Choice! Sure there is an inductive algorithm, since we can assume that we may pick an element from a given set: pick $x_0$ from $A_0$, pick $x_1$ from $A_1$ ... This is entirely unnecessary though – basket Jul 01 '16 at 09:14
  • Have you read the infinitely many threads written on this before? – Asaf Karagila Jul 01 '16 at 09:18
  • @AsafKaragila they looked a bit different, or I missed some? – mnaoumov Jul 01 '16 at 09:24
  • For example, http://math.stackexchange.com/questions/85153/how-do-i-choose-an-element-from-a-non-empty-set – Asaf Karagila Jul 01 '16 at 10:23
  • @AsafKaragila thanks, I found that question interesting, and actually I can rephrase my question now – mnaoumov Jul 01 '16 at 11:41
  • @CameronBuie clearly it is a duplicate. I overlooked it! – mnaoumov Jul 01 '16 at 11:58

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