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Suppose $\kappa$ is an uncountable cardinal. Then $S\subseteq\kappa$ is club (CLosed UnBounded) if $S$ is unbounded in $\kappa$ and is a closed subset of $\kappa$ under the order topology. My question is the following: suppose $S\subseteq\kappa$ is club, and $S=S_1\cup S_2$. Is there necessarily some club set $C$ such that either $C\subseteq S_1$ or $C\subseteq S_2$?

(Note that, since the intersection of two clubs is again club, this question amounts to asking whether, for $\kappa$ an infinite cardinal, the set $\{S\subseteq\kappa: S$ is a superset of some club set $\}$ is an ultrafilter.)

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up vote 3 down vote accepted

The answer is no, this is never an ultrafilter. This uses choice. We assume that the cofinality of $\kappa$ is bigger than $\omega$, else, $C$ has a subset of type $\omega$ cofinal (which is automatically club), and this can be split into two disjoint cofinal (therefore club) sets.

If the cofinality of $\kappa$ is larger than $\omega_1$, simply note that for any regular $\alpha<\kappa$, $\{\beta<\kappa:\beta$ has cofinality $\alpha\}$ is stationary, i.e., it meets every club (consider the $\alpha$-th member of the increasing enumeration of the club. These sets are disjoint for different $\alpha$, and if $\kappa>\omega_1$, there are at least two regular $\alpha$ below $\kappa$, namely $\omega$ and $\omega_1$. This shows that we can split $C$ into disjoint stationary sets (and therefore, neither is club, since a stat. set must meet every club).

All I've used so far is that $\omega_1$ is regular.

To show that clubs on ordinals of cofinality $\omega_1$ can be split into disjoint stationary sets requires more choice. In fact, it is consistent that the club filter on $\omega_1$ is an ultrafilter. This is the case, for example, in models where the axiom of determinacy holds. (But it is strictly weaker than determinacy. A measurable suffices in consistency strength.)

The typical argument from choice uses the existence of Ulam matrices. See definition 12 and the following paragraph in this blog entry. An $\omega\times\omega_1$ Ulam matrix gives us that every stat. subset of $\omega_1$, in particular every club, can be split into $\omega_1$ disjoint stat. sets. So the club filter is not an ultrafilter, and we are done.

Ulam's argument also shows that any stat. subset of $\kappa^+$ can be split into $\kappa^+$ stat. sets, and Solovay proved the same for any reglar $\lambda$, not just a successor. This, of course, gives the result that no club filter is an ultrafilter, but I wanted to point out that only in cofinality $\omega_1$ we need that much choice to carry out the argument.

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This is an excellent answer! Thank you very much. – user5064 Dec 23 '10 at 7:42
It might be worth adding that the consistency of "The club filter is an ultrafilter on $\omega_1$" is indeed a measurable, but it's more than just one measurable. Much much more than just one measurable. – Asaf Karagila Nov 30 '14 at 20:19
(Although we can have the club filter on $\omega_1$ as an ultrafilter without any large cardinals; but without countable choice either, and it won't be $\sigma$-complete either.) – Asaf Karagila Nov 30 '14 at 20:29

If you are familiar with the idea of clubs and stationary, then there is a good measure theoretic intuition around it:

Clubs are of measure $1$. They are almost everywhere.
Non-stationary sets are of measure $0$ and are almost nowhere.
Stationary sets are of positive measure.

Now to your question, look at the unit interval, and take some subset of measure $1$, you can split it into two positive measured sets, none of which is of measure $1$, namely intersect it with $[0,\frac{1}{2})$ and with $[\frac{1}{2},1]$, and you have two disjoint subsets of positive measure.

Now it is important to say that this is just the intuition at heart, and since Andres gave a good answer with a formal backbone I decided to add this one as well.

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