# Does “$|{\cal P}_2(X)| = |X|$ for $X$ infinite” imply ${\sf (AC)}$?

This comes from a comment made by user bof in this thread.

Let $X$ be a set, define ${\cal P}_2(X) = \big\{\{a, b\}: a\neq b\in X\big\}$.

Consider the statement

${\sf (S)}$ If $X$ is an infinite set, then there is a bijection $\varphi: {\cal P}_2(X)\to X$.

Does ${\sf (S)}$ imply ${\sf (AC)}$?

## 1 Answer

Yes, the usual proof that $|X|^2=|X|$ for all $X$ implies AC works for (S) as well. In detail, let $A$ be any infinite set, let $H$ be its Hartogs number (the least ordinal that does not inject into $A$), and let $X=A\sqcup H$. A bijection $\mathcal{P}_2(X)\to X$ in particular restricts to an injection $i:A\times H\to X$. For each $a\in A$, by definition of $H$ there must be some $h\in H$ such that $i(a,h)\not\in A$. Define $j:A\to H$ by $j(a)=i(a,h)$ for the least such $h$. Then $j$ is an injection. Since $H$ is well-orderable, it follows that $A$ is well-orderable.