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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?

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The other direction is trickier because it requires more assumptions. Specifically, the implication from "The power set of an ordinal can be well-ordered" to "The axiom of choice" requires the use of the axiom of foundation. Which is usually less natural when it comes to thinking about mathematics... naively.

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.

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  • $\begingroup$ What is meant by $V_{\alpha}$? $\endgroup$ – Oiler May 2 '16 at 6:46
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila May 2 '16 at 6:47
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    $\begingroup$ Just clarifying a notation. The last part is unnecessary. $\endgroup$ – Oiler May 2 '16 at 6:57
  • $\begingroup$ I wasn't trying to be offensive. $\endgroup$ – Asaf Karagila May 2 '16 at 7:09

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