TITLE: For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
QUESTION [50 upvotes]: How to prove the following conclusion:
[For any infinite set $S$, there exists a bijection $f:S\to S \times S$] implies the Axiom of choice.
Can you give a proof without the theory of ordinal numbers.

REPLY [27 votes]: Let $A$ be an arbitrary infinite set. I will show that $A$ has a choice function, i.e., a function $a:\mathcal{P}(A)\setminus\{\varnothing\}\to A$ such that $a(X) \in X$ for every nonempty $X \subseteq A$.
First, pick a set $B$ such that:

there is a choice function $b:\mathcal{P}(B)\setminus\{\varnothing\}\to B$, and

there is no injection from $B$ into $A$.


(For example, take $B$ to be the set of all isomorphism classes of wellorderings of subsets of $A$.) Assume further that $A \cap B = \varnothing$.
By hypothesis, there is an injection $f:(A\cup B)^2\to(A\cup B)$. Given $x \in A$, there must be a $y \in B$ such that $f(x,y) \in B$, otherwise $y \in B \mapsto f(x,y)$ would be an injection from $B$ into $A$. We may thus define a function $g:A \to B^2$ by $g(x) = (y,f(x,y))$ where $y = b(\{u \in B : f(x,u) \in B\})$, i.e., for each $x \in A$, $g(x)$ picks a pair $(y,z) \in B^2$ such that $f(x,y) = z$. Note that $g$ must be an injection: if $g(x) = (y,z) = g(x')$ then $f(x,y) = z = f(x',y)$ and hence $x = x'$ since $f$ is an injection.
Now observe that $B^2$ has a choice function just like $B$. Namely, let $c:\mathcal{P}(B^2)\setminus\{\varnothing\}\to B^2$ be defined by $c(X) = (y,z)$ where $$y = b(\{u \in B : (\exists v \in B)((u,v) \in X)\})$$ and $$z = b(\{v \in B : (y,v) \in X \})$$ for every nonempty $X \subseteq B^2$.
It follows that $A$ has a choice function too. Namely, let $a:\mathcal{P}(A)\setminus\{\varnothing\}\to A$ be defined by $$a(X) = g^{-1}(c(\{g(x) : x \in X\}))$$ for every nonempty $X \subseteq A$.
Remark. The proof of the existence of the set $B$ outlined above relies on the theory of wellorderings but not on the theory of ordinals. In fact, the above argument can be carried out in the theory Z (= ZF minus the Replacement and Foundation axioms). It is well known that there are models of Z with very few ordinals, e.g., $V_{\omega+\omega}$ is a model of Z.