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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.

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In order for the clubs to form a filter, you need $\alpha$ to have uncountable cofinality. –  Amit Kumar Gupta Dec 27 '10 at 11:35
    
Also if choice fails, a countable union of countable sets might not even be countable I believe, and so the notion of "closed" might not even make sense in the generalized setting. –  Amit Kumar Gupta Dec 27 '10 at 11:38
    
You may also want to tag this question with set-theory, axiom-of-choice, and ultrafilters to increase visibility. –  Jason Dec 28 '10 at 4:39
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2 Answers

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.

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Looking further in the Cabal seminars, I see in the 1979-81 volume (SLNM 1019) a paper by Woodin, "AD and the uniqueness of the supercompact measures on $P_{\omega_1}(\lambda$," whose Theorem 4 says, under the hypothesis $AD_{\mathbb R}$, that if $S$ is a surjective image of $\mathbb R$ than the set of countable subsets of $S$ carries a unique supercompact measure. Unfortunately, I again don't see anything about whether this is the club filter. –  Andreas Blass Dec 28 '10 at 3:08
    
Andreas, the result actually holds in AD${}^+$. This is not entirely straightforward; more or less, one needs to translate the real games of Woodin's paper using the generic-coding arguments introduced in the paper by Kechris and Woodin (that appeared in the new Cabal volume). Anyway, the results in the paper you mention show that the supercompactness measures are precisely the club filters. (Actually, they are usually called "weak club filters", to emphasize that we only require closure under (increasing) countable unions. –  Andres Caicedo Dec 28 '10 at 6:48
    
Hmm... Let me try again: $S\subseteq{\mathcal P}_{\omega_1}(\kappa)$ is a weak club set iff $\bigcup S=\kappa$ and whenever $\sigma_0\subseteq\sigma_1\subseteq\dots$ and each $\sigma_i\in S$, then $\bigcup_i\sigma_i\in S$. Woodin's result shows that (under AD${}_{\mathbb R}$ or AD${}^+$) if $\kappa$ is a Suslin cardinal, then there is a unique normal fine measure on ${\mathcal P}_{\omega_1}(\kappa)$, and it coincides with the weak club filter. For $\gamma<\kappa$, the unique such measure on ${\mathcal P}_{\omega_1}(\gamma)$ is obtained by projecting. –  Andres Caicedo Dec 28 '10 at 7:40
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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 $\omega_1$.

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