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?


1 Answer 1


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 $f|\alpha R f(\alpha)$.

  • $\begingroup$ I literally snoozed and missed answering the question. :-) $\endgroup$
    – Asaf Karagila
    Apr 8, 2015 at 20:39
  • 1
    $\begingroup$ And I used the opportunity, all the while wondering why you weren't beating me to it :). $\endgroup$ Apr 8, 2015 at 21:22
  • $\begingroup$ It was 37 degrees centigrade today. Tomorrow is 21 degrees. The weather is crazy. I also have to grade homework assignments. All these things drained my will to stay awake... :-) $\endgroup$
    – Asaf Karagila
    Apr 8, 2015 at 21:23
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Apr 8, 2015 at 21:30
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    $\begingroup$ $\sf DC_\kappa\implies AC_\kappa, W_\kappa$ and that is the only implication, without adding more assumptions. $\endgroup$
    – Asaf Karagila
    Apr 8, 2015 at 21:33

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