1
$\begingroup$

This question already has an answer here:

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 $A$?

$\endgroup$

marked as duplicate by Cameron Buie, Community Jul 1 '16 at 12:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If a set of certain objects exists then certainly one such object exists, yes? $\endgroup$ – basket Jul 1 '16 at 9:05
  • $\begingroup$ @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 $\endgroup$ – mnaoumov Jul 1 '16 at 9:13
  • $\begingroup$ 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 $\endgroup$ – basket Jul 1 '16 at 9:14
  • $\begingroup$ Have you read the infinitely many threads written on this before? $\endgroup$ – Asaf Karagila Jul 1 '16 at 9:18
  • $\begingroup$ @AsafKaragila they looked a bit different, or I missed some? $\endgroup$ – mnaoumov Jul 1 '16 at 9:24

Browse other questions tagged or ask your own question.