Consistency strength of the "club ultrafilter"

I should have corrected this years ago. There is a mistake in Spector's argument outlined in the paper (the equiconsistent one, that is).

Enumerate all the stationary subsets of $\omega_1$ in the ground model. By the very nature of the lottery iteration, we will have picked arbitrarily high in this enumeration, which means that we collapsed, at least $\omega_1$ itself. The problem, of course, is that the lottery sum of homogeneous forcings need not be homogeneous itself, so the argument that any set of ordinals was added in a bounded stage of the iteration does not work.

(This was observed independently by some other people as well.)

Anyway, to recover this proof one can try to somehow deal with all ground model sets at once. So, for example, choose an ultrafilter extending the club filter (or add a generic one), and shoot clubs through those sets. Now rinse and repeat.

But the problem now is that the choice of the ultrafilter (or the forcing adding it) makes the whole thing very non-homogeneous, so we face the same issue as above. This was one of the few routes I've tried with Yair Hayut, but of course, we may have missed something.

So to my best understanding, it is still entirely open if you can have a club ultrafilter from "weak assumptions" (i.e. weaker than Mitchell's known bounds).

The first statement is equiconsistent with $\sf ZF$, without large cardinals. The proof is due to Spector,

Mitchell Spector, The $\kappa $-closed unbounded filter and supercompact cardinals, J. Symbolic Logic 46 (1981), no. 1, 31--40.

The outline of the model is not hard. Consider $\Bbb P$ to be the forcing which is the lottery sum of all forcings which add a club into a stationary $S$ which is co-stationary. Now iterate $\Bbb P$ with itself $\omega$ times with finite support (each step re-calculating $\Bbb P$, of course), and consider the model definable by bounded steps of the iteration.

Since we didn't collapse $\omega_1$ at any bounded step, it remains regular in this model, and we can show that any stationary set in this model was added at a bounded step, so by genericity was chosen by the lottery sum to have its complement become non-stationary, so it contains a club.

Interestingly, the intersection of $\omega$ clubs is still a club, but given $\omega$ sets, you can't choose a club subset of each one uniformly. So the filter is not $\sigma$-closed because of that.

Once you want to assume $\sf DC$ into the mix, the club filter must become a $\sigma$-closed filter, so we need more. At least a measurable, but in fact more. In the above paper Spector proves this from a supercompact cardinal.

Mitchell and Martin show that this requires more than just one measurable cardinal, by showing that for each $\delta<\omega_1$ this statement implies an inner model with $\delta$ measurable cardinals.

D. A. Martin and W. Mitchell, On the ultrafilter of closed, unbounded sets, J. Symbolic Logic 44 (1979), no. 4, 503--506.

Mitchell himself gave an upper bound of $\kappa$ measurable with $o(\kappa)=\kappa^{++}$,

Mitchell, William, How weak is a closed unbounded ultrafilter?, Logic Colloquium '80 (Prague, 1980), pp. 209–230. (MR673794)

I was told that this was later improved to $o(\kappa)=\kappa^+$, but I cannot find a reference at this time.

It might be worth mentioning that the consistency of there exists a countably complete measure on some ordinal $\kappa$ is indeed a measurable. We can even get it to be $\omega_1$, as Jech showed in his paper,

T. Jech, $\omega _{1}$ can be measurable, Israel J. Math. 6 (1968), 363--367 (1969).

But there is still a great distance between just a measure on $\omega_1$, to having that measure as the club filter.

Perhaps this would be a good place to point out that it is equiconsistent with $\sf ZFC$ that there is an infinite set with a countably complete ultrafilter. Moreover we can have this ultrafilter be the cofinite filter. We can also have, for no additional consistency strength a countably closed ultrafilter on a set with $\sf ZF+DC$, and again we can arrange for it to be the cocountable filter.

This process can continue, but it should be pointed out that these sets are not well-orderable at all. And this makes using their ultrafilters for "usual purposes" a bit irrelevant.