# Consistency strength of the “club ultrafilter”

What are the consistency strengths of $$ZF+\text{The club filter on \omega_1 is an ultrafilter}"$$ and $$ZF + DC + \text{The club filter on \omega_1 is an ultrafilter}"?$$

I know that the latter (stronger) statement follows from AD, and has consistency strength much weaker than AD (I think a measurable is enough), but I don't know what the precise strength of either statement is. In particular, I suspect that the latter has strictly greater consistency strength than the former, but I don't know if this is the case.

EDIT: of course, if the club filter is an ultrafilter, then it is countably complete (since it is countably complete as a filter), so $\omega_1$ is measurable; so it may appear that the answer to this question is trivially "a measurable." However, this isn't quite right: it is not obvious (in fact, I suspect it's not true) that $ZF+$"there is a countably complete ultrafilter on some cardinal $\kappa$" and $ZFC+$"there is a countably complete ultrafilter on some cardinal $\kappa$" are equiconsistent.

FURTHER EDIT: As Asaf shows below, the club filter need not be countably closed in $ZF$, even though the intersection of countably many clubs is always club!

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.

• Am I right in thinking that it's inconsistent to assume that $\omega_1$ is the critical point of a $j:V\prec M$? If so, where does the usual proof that $\omega_1$'s having a countably complete (non-principle) ultrafilter is equivalent to the embedding condition fail in ZF + DC? Thanks in advance! – GME Nov 6 '14 at 10:02
• Well, that's in general more complicated than that. If $W$ is a model with a measurable $\kappa$ and $V$ is the generic extension of $W$ by collapsing every ordinal below $\kappa$ to be countable (so $\kappa=\omega_1^V$), then there is a generic extension $V[G]$ in which there is an elementary embedding $V\to M$ with critical point $\kappa=\omega_1^V$. We can even have $M$ as the generic extension of $W^\kappa/U$ by collapsing all the ordinals below $j(\kappa)$ to be countable (where $U\in W$ is a normal measure on $\kappa$). – Asaf Karagila Nov 6 '14 at 10:08
• That's cool. But, just to be clear, $\kappa$ is not $\omega_1$ in $V[G]$ right? Because that would be inconsistent? Here's another way to put my question: can there be a model of ZF + DC + "there is a countably complete ultrafilter on $\omega_1$"? – GME Nov 6 '14 at 10:15
• You're right, $\kappa$ is not $\omega_1$ in $V[G]$. But it is still the critical point etc. etc.; to you "another way" question, yes $\sf ZF+DC+\omega_1\text{ is measurable}$ is equiconsistent with $\sf ZFC+\exists\kappa\text{ measurable}$. That's what Jech showed, and that's what we get in $L(\Bbb R)$ when there are sufficiently many Woodin cardinals. – Asaf Karagila Nov 6 '14 at 10:26
• So that means $\kappa$ will be countable in $V[G]$ (right?). That's really cool! I thought Jech just showed the consistency of ZF + $\omega_1$ is measurable. Is it simple to see how to extend his result to get a model of DC also? Or a reference? Thanks again! – GME Nov 6 '14 at 10:34