[SOLVED] How strong is the axiom of well-ordered choice?

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?

@Asaf Karagila 2019-03-08 22:15:19

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.

@Holo 2019-03-08 23:14:33

Do you have a source for a proof of ${\sf AC_{\rm WO}}\iff \forall x(\aleph(x)=\aleph^*(x))$?

@Asaf Karagila 2019-03-09 08:52:57

(The proof there assumes PP, but really you only use that for maps onto ordinals, which is the same as saying $\aleph=\aleph^*$.)

@Holo 2019-03-09 16:43:14

thank you very much!