# Logical strength of "choice functions exist for well-ordered families"?

A colleague of mine suggested the following weakening of the axiom of choice:

If $\mathscr{F} := \{F_\alpha\}$ is a well-ordered family of non-empty sets (i.e., there is a bijection between $\mathscr{F}$ and some von Neumann ordinal), then there is a choice function $f$ with domain $\mathscr F$ such that $f(F_\alpha) \in F_\alpha$ for all $\alpha$.

The intuition behind this axiom is that the well ordering allows us to imagine picking the representatives "one at a time," and at any stage there is always a clear candidate for the next $F_\alpha$ to consider.

Question: How does this axiom compare to other known weak forms of the axiom of choice?

Clearly it implies countable choice, but does it imply anything interesting that countable choice does not?

• I actually outlined a proof of what Vika wrote in her answer, in a question she had asked some time ago. mathoverflow.net/questions/178473/… Apr 8, 2015 at 21:25

This principle is known in set-theoretic literature as $\forall \kappa AC_\kappa$, where $AC_\kappa$ refers to the existence of choice functions for well-ordered families of size $\kappa$. It is discussed in detail in Chapter 8 of Jech's textbook "The axiom of choice". The principle $\forall\kappa AC_\kappa$ implies slightly more than the countable axiom of choice, namely it implies $DC_\omega$, the axiom of dependent choices, which states that any set relation without terminal nodes has an infinite branch. But $\forall \kappa AC_\kappa$ does not imply the generalization of $DC_\omega$ to $\omega_1$, $DC_{\omega_1}$, which allows $\omega_1$-many dependent choices. On the other hand, the principle $\forall \kappa DC_\kappa$ does turn out to be equivalent to $AC$. More precisely, $DC_\kappa$ states that for any set $S$ and any relation $R$ on $S^{<\kappa}\times S$, if for every $\vec x$ in $S^{<\kappa}$, there is $x\in S$ such that $\vec x R x$, then there is $f:\kappa\to S$ such that $f|\alpha R f(\alpha)$.
• Well, only the minor addition that it does not imply that every set is comparable with $\aleph_1$ in cardinality (which is weaker than $\sf DC_{\omega_1}$ of course). I actually linked to your question regarding these proofs from the summer, below in the comments of the question. Apr 8, 2015 at 21:30
• $\sf DC_\kappa\implies AC_\kappa, W_\kappa$ and that is the only implication, without adding more assumptions. Apr 8, 2015 at 21:33