Generalization of the club filter

Question

If $\alpha$ is a (limit) ordinal, then a subset $S\subseteq\alpha$ is club if $\alpha$ is closed as a subset of $\alpha$ under the order topology and unbounded in $\alpha$. The set of all sets containing a club forms a filter on the subsets of $\alpha$, called the club filter.

This definition can be extended in the following way. Let $C$ be any infinite set, and let $P$ be the set of all countable subsets of $C$. Then $S\subseteq P$ is club if $S$ is closed under unions of countable chains (closed), and for all $X\in P$, there is some $Y\in S$ with $X\subseteq Y$ (unbounded).

(One area where this notion of clubness appears is in proofs of a Lowenheim-Skolem theorem for infinitary logic; see e.g. http://www.math.uic.edu/~marker/dwk.pdf.)

Given a set $C$, we can take the set of all subsets of $P$ which contain a club (in the generalized sense); call this the club filter on $P$.

My question is: when is the club filter in this latter case an ultrafilter?

If I understand things correctly, there should be no possibility of the club filter being an ultrafilter if we assume AC. However, in the original sense of the word club, the club filter on $\omega_1$ is an ultrafilter assuming AD, so this leads me to believe that, in ZF + AD, there might be interesting sets $C$ the club filter of which is an ultrafilter.

In particular, what kinds of choice need to fail at $C$ or $P$ in order for the club filter on $P$ to be an ultrafilter?

I hope this question is meaningful; I don't have much background knowledge of models of set theory in which choice fails.

Answer 1

Here's a little piece of an answer: Section 3 of Solovay's paper "The independence of DC from AD" (Cabal Seminar 76-77, Springer Lecture Notes in Math 689 (1978) pp. 171-183) has a construction of a normal, countably complete ultrafilter $U$ on the set of countable subsets of the real line. The construction assumes $AD_{\mathbb R}$, the axiom of determinacy for games in which each move is a real number. The definition of $U$ is that it contains those sets $A$ (of countable sets of reals) for which player II has a winning strategy in the following game. The players alternately choose finite sets of reals, and II wins a play iff the union of the chosen sets is an element of $A$. Looking through the paper rather quickly, I didn't find anything saying whether this $U$ is the club filter, and it's too late in the evening for me to figure it out, but it looks to me as if it has a reasonable chance of being the club filter.

Answer 2

First, in order to avoid Amit's concern, we may as well assume countable choice. Note that if AD (axiom of determinacy) holds in $L(\mathbb{R})$, then DC (axiom of dependent choice) will also be true there, and DC implies countable choice.

What you are asking for then is related to Jech's notion of stationarity. Specifically, we say that $S \subseteq [A]^{\omega}$ ($[A]^{\omega}$ is the set of countable subsets of $A$) is stationary when $S$ meets every club, where club here is in the sense you described. Now you can verify that the club filter on $[A]^{\omega}$ is an ultrafilter if and only if $[A]^{\omega}$ cannot be decomposed into two disjoint stationary sets.

Although not true for arbitrary $A$, you can also verify that $S \subseteq [\omega_1]^{\omega}$ is stationary according to Jech's characterization if and only if $S \cap \omega_1$ is stationary in the usual sense. Since AD implies that the club filter on $\omega_1$ is an ultrafilter ( Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? ), I claim that the club filter on $[\omega_1]^{\omega}$ is actually an ultrafilter if AD is true for suppose not. Then the club filter on $[\omega_1]^{\omega}$ could be decomposed into two disjoint stationary sets $S_1$ and $S_2 = [\omega_1]^{\omega} \setminus S_1$ so that $S_1 \cap \omega_1$ and $S_2 \cap \omega_1$ would be disjoint stationary sets of $\omega_1$. But this is impossible because then $S_1 \cap \omega_1$ and its complement would not be in the club filter on $\omega_1$.