I sometimes see references to the "Axiom of Well-Ordered Choice," but I'm not sure how strong it is. It states that every well-ordered family of sets has a choice function.

By "well-ordered family," I don't mean that the sets within the family are well-ordered, but that the family must index all the sets within the family by some ordinal.

How strong is this axiom? Can it prove the Hahn-Banach theorem, the ultrafilter lemma, anything about measurable sets, etc? Does it have any implications about what sets can be well-ordered (the reals for instance), or perhaps prove anything about the Hartogs number of sets, etc?

Does anyone have a reference for this?

  • $\begingroup$ ...isn't this axiom a consequence of ZF? One can choose the (well-defined, since it's well-ordered) lexicographically least element of each set in the family... $\endgroup$ Mar 8, 2019 at 20:46
  • 2
    $\begingroup$ That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets). $\endgroup$ Mar 8, 2019 at 20:47
  • $\begingroup$ Ahh, I missed that distinction. Thank you! $\endgroup$ Mar 8, 2019 at 20:50
  • $\begingroup$ I've never seen this axiom before. Does the family itself need to be a set or can it be a proper class? $\endgroup$ Mar 8, 2019 at 20:55
  • $\begingroup$ Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice $\endgroup$ Mar 8, 2019 at 20:56

1 Answer 1


The axiom of well-ordered choice, or $\sf AC_{\rm WO}$, is strictly weaker than the axiom of choice itself. If we start with $L$ and add $\omega_1$ Cohen reals, then go to $L(\Bbb R)$, one can show that $\sf AC_{\rm WO}$ holds, while $\Bbb R$ cannot be well-ordered there.

Pincus proved in the 1970s that this is equivalent to the following statement on Hartogs and Lindenbaum numbers:

$\sf AC_{\rm WO}$ is equivalent to the statement $\forall x.\aleph(x)=\aleph^*(x)$.

Here, the Lindenbaum number, $\aleph^*(x)$, is the least ordinal which $x$ cannot be mapped onto. One obvious fact is that $\aleph(x)\leq\aleph^*(x)$.

In the late 1950s or early 1960s Jensen proved that this assumption also implies $\sf DC$. This is also a very clever proof.

The conjunction of these two consequences gives us that $\aleph_1\leq 2^{\aleph_0}$, as a result of a theorem of Shelah from the 1980s, this implies there is a non-measurable set of reals.

As far as Hahn–Banach, or other things of that sort, I do not believe that much is known on the topic. But to sum up, this axiom does not imply that the reals are well-ordered, but it does imply there is a non-measurable set of reals because there is a set of reals of size $\aleph_1$ and $\sf DC$ holds. Moreover, it is equivalent to saying that the Hartogs and Lindenbaum numbers are equal for all sets.

  • $\begingroup$ Do you have a source for a proof of ${\sf AC_{\rm WO}}\iff \forall x(\aleph(x)=\aleph^*(x))$? $\endgroup$
    – ℋolo
    Mar 8, 2019 at 23:14
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    $\begingroup$ @Holo karagila.org/2014/on-the-partition-principle $\endgroup$
    – Asaf Karagila
    Mar 8, 2019 at 23:26
  • $\begingroup$ (The proof there assumes PP, but really you only use that for maps onto ordinals, which is the same as saying $\aleph=\aleph^*$.) $\endgroup$
    – Asaf Karagila
    Mar 9, 2019 at 8:52
  • $\begingroup$ thank you very much! $\endgroup$
    – ℋolo
    Mar 9, 2019 at 16:43

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