Without Foundation, can axiom of choice be derived from $\forall\alpha\in\mathbf{Ord} (P(\alpha)\text{ can be well ordered})$?

I don't think it works with just ZF minus foundation ($\mathbf{ZF}^-$). To get your choice function (which you only know to work in the wellfounded part of the universe) to work in illfounded part, you would require a version of Coret's axiom

Every set has the same size as a wellfounded set

(or something similar) which is not a theorem of $\mathbf{ZF}^-$.


It is impossible to prove this implication without the axiom of regularity. The proof is given in Lemma 9.4 of Jech's Axiom of choice.

Jech uses the ordered Mostowski model to separate the well-ordering principle for $\mathcal{P}(\alpha)$ and that every linearly orderable set is well-orderable (which is a consequence of the axiom of choice.) However, taking any permutation model works to separate the well-ordering principle for $\mathcal{P}(\alpha)$ and the axiom of choice, as Jech observed in his book.

Here it is why: If $\alpha$ is an ordinal, then every subset of $\mathcal{P}(\alpha)\times \mathcal{P}(\alpha)$ is hereditarily symmetric (since it contains no atoms.) Especially, every well-order of $\mathcal{P}(\alpha)$ is still hereditarily symmetric. (Note that we assume the axiom of choice for the ground model. The point is that we do not know there is a hereditarily symmetric well-order of any given set.)

As a corollary, we may have an amorphous set or Russell socks set, while every power set of an ordinal is still well-orderable.

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Set Theory

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