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Define: $\operatorname {wo}^n(x) \iff \forall y (y \in^n x \to \operatorname {wo} (y))$

Where $\operatorname {wo}(y)$ refers to $y$ being well orderable.

Where $y \in^0 x \iff y=x \\ y \in^{n+1} x \iff \exists z (z \in^n x \land y \in z)$

n-well ordered choice: for $n=0,1,2,...$, for every set $x$ of nonempty sets, if $\operatorname{wo}^n(x)$, then $x$ admits a choice function on it.

If we add this schema to axioms of $\sf ZF$, would it entail axiom of choice?

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$2$-well ordered choice is enough to imply AC.

Let $α$ be any ordinal, and look at $\mathcal P^2(α)\setminus\{\emptyset\}$.

We have $x\in^2 \mathcal P^2(α)\setminus\{\emptyset\}⇒∃z⊆\mathcal P(α)\;(x\in z)⇒x\subseteq α⇒x\text{ is well orderable}$.

A choice function on $\mathcal P^2(α)\setminus\{\emptyset\}$ induce a well ordering on $\mathcal P(α)$.

"powerset of well-orderable set is well-orderable" is famously equivalent to the axiom of choice.

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  • $\begingroup$ what is $\mathcal P^2$? $\endgroup$
    – Zuhair Al-Johar
    Commented Feb 22, 2023 at 21:54
  • $\begingroup$ @ZuhairAl-Johar the powerset of the powerset $\endgroup$
    – Holo
    Commented Feb 22, 2023 at 21:56
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No. In the Cohen model every family of well orderable sets admits a choice function. So $1$-well ordered choice holds, but countable choice fails.

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  • $\begingroup$ This is an answer to $1$-well orderable choice. While with the schema I've written we have an axiom for each $n$. You need to show a model that fulfills all of those axioms not just only $1$-well orderable choice, and in which choice fails. So, this doesn't answer this particular question. $\endgroup$
    – Zuhair Al-Johar
    Commented Feb 22, 2023 at 21:04

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