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?
Best Answer
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:
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.