# Proving the Axiom of Choice is equalivalent to the statement “If $A$ can be well-ordered, then so can $\mathcal{P}(A)$.”

I am still not completely confident in proving equivalence between the Axiom of Choice and statements such as the one posed in the title, so I want to make sure that I am on the right track.

So showing these two statements are equivalent requires me to show that they each imply each other. So if I assume the Axiom of Choice, then certainly $\mathcal{P}(A)$ can be well-ordered because we know the Axiom of Choice is equivalent to the statement "every set can be well-ordered."

The other direction is a little murkier for me. Would assuming the statement and then exhibiting a choice function on $\mathcal{P}(A)$ imply the Axiom of Choice and affirm these two statements are equivalent?

We also know this is necessary. In $\sf ZFA$, which is equivalent to some variant of $\sf ZF-Fnd$, this implication is false.
The general idea is to prove, by induction that every $V_\alpha$ can be well-ordered. Of course the trick is at limit steps, how to coalesce all the well-orders of the previous steps into a well-ordering of $V_\delta$ without having to use the axiom of choice.
• What is meant by $V_{\alpha}$? – Oiler May 2 '16 at 6:46
• The $\alpha$th level of the von Neumann hierarchy. If you're not clear what that is, then you're underprepared to tackle this problem. – Asaf Karagila May 2 '16 at 6:47