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Let

$\mathsf{AC}_\mathsf{WO}$: Every well-orderable family of non-empty sets has a choice function.

$\mathsf{AC}^\mathsf{WO}$: Every family of non-empty well-orderable sets has a choice function.

My questions are:

  • Does $\mathsf{ZF}+\mathsf{AC}_\mathsf{WO}+\mathsf{AC}^\mathsf{WO}$ prove $\mathsf{DC}_{\omega_1}$?
  • Does $\mathsf{ZF}+\mathsf{AC}_\mathsf{WO}+\mathsf{AC}^\mathsf{WO}$ prove $\mathsf{AC}$?

I've searched in Howard and Rubin's Consequences of the Axiom of Choice without success. For sure both $\mathsf{AC}_\mathsf{WO}$ and $\mathsf{AC}^\mathsf{WO}$ do not imply $\mathsf{AC}$, more specifically $\mathsf{AC}_\mathsf{WO}$ doesn't prove $\mathsf{DC}_{\omega_1}$ and $\mathsf{AC}^\mathsf{WO}$ doesn't even prove $\mathsf{AC}_\omega(\mathbb{R})$.
Do you have an idea?

EDIT: In Howard, Paul; Rubin, Jean E., The axiom of choice for well-ordered families and for families of well-orderable sets, it is proven that we cannot find a Fraenkel–Mostowski model witnessing such an independence. According to the authors, it is an open problem (unless some progress has been made in the mean time).

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    $\begingroup$ I don't have an example or proof off hand, but I'd be very surprised if the answer is positive. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 21, 2022 at 16:18
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    $\begingroup$ @AsafKaragila Me too! But all the symmetric models I've seen in which $\mathsf{AC}^{WO}$ holds use finite supports and the least support property, whilst to have $\mathsf{AC}_{WO}$ we cannot use finite supports I think, at least in the usual à la Cohen way... $\endgroup$
    – Lorenzo
    Commented Oct 21, 2022 at 16:21
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    $\begingroup$ MathJax/LaTeX tip: $\mathsf{ZF}+\mathsf{AC}_\mathsf{WO}+\mathsf{AC}^\mathsf{WO}$ looks exactly the same as $\mathsf{ZF+AC_{WO}+AC^{WO}}$. $\endgroup$
    – Emil Jeřábek
    Commented Oct 22, 2022 at 20:16
  • $\begingroup$ Can you perhaps think of a single statement of the form “every <foo> family of non-empty <bar> sets” which would imply both $\mathsf{AC_{WO}}$ and $\mathsf{AC^{WO}}$ without obviously implying $\mathsf{AC}$ (even if non-implication isn't clear either)? That might be a start. $\endgroup$
    – Gro-Tsen
    Commented Oct 22, 2022 at 22:06
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    $\begingroup$ The problem with FM models here is that well-orderable is too strict of a notion, in a sense. Specifically, the power set of a well-orderable set is again well-orderable. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 23, 2022 at 9:23

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