Club filter

In mathematics, particularly in set theory, if $$\kappa$$ is a regular uncountable cardinal then $$\operatorname{club}(\kappa),$$ the filter of all sets containing a club subset of $$\kappa,$$ is a $$\kappa$$-complete filter closed under diagonal intersection called the club filter.

To see that this is a filter, note that $$\kappa \in \operatorname{club}(\kappa)$$ since it is thus both closed and unbounded (see club set). If $$x\in\operatorname{club}(\kappa)$$ then any subset of $$\kappa$$ containing $$x$$ is also in $$\operatorname{club}(\kappa),$$ since $$x,$$ and therefore anything containing it, contains a club set.

It is a $$\kappa$$-complete filter because the intersection of fewer than $$\kappa$$ club sets is a club set. To see this, suppose $$\langle C_i\rangle_{i<\alpha}$$ is a sequence of club sets where $$\alpha < \kappa.$$ Obviously $$C = \bigcap C_i$$ is closed, since any sequence which appears in $$C$$ appears in every $$C_i,$$ and therefore its limit is also in every $$C_i.$$ To show that it is unbounded, take some $$\beta < \kappa.$$ Let $$\langle \beta_{1,i}\rangle$$ be an increasing sequence with $$\beta_{1,1} > \beta$$ and $$\beta_{1,i} \in C_i$$ for every $$i < \alpha.$$ Such a sequence can be constructed, since every $$C_i$$ is unbounded. Since $$\alpha < \kappa$$ and $$\kappa$$ is regular, the limit of this sequence is less than $$\kappa.$$ We call it $$\beta_2,$$ and define a new sequence $$\langle\beta_{2,i}\rangle$$ similar to the previous sequence. We can repeat this process, getting a sequence of sequences $$\langle\beta_{j,i}\rangle$$ where each element of a sequence is greater than every member of the previous sequences. Then for each $$i < \alpha,$$ $$\langle\beta_{j,i}\rangle$$ is an increasing sequence contained in $$C_i,$$ and all these sequences have the same limit (the limit of $$\langle\beta_{j,i}\rangle$$). This limit is then contained in every $$C_i,$$ and therefore $$C,$$ and is greater than $$\beta.$$

To see that $$\operatorname{club}(\kappa)$$ is closed under diagonal intersection, let $$\langle C_i\rangle,$$ $$i < \kappa$$ be a sequence of club sets, and let $$C = \Delta_{i<\kappa} C_i.$$ To show $$C$$ is closed, suppose $$S\subseteq \alpha < \kappa$$ and $$\bigcup S = \alpha.$$ Then for each $$\gamma \in S,$$ $$\gamma \in C_\beta$$ for all $$\beta < \gamma.$$ Since each $$C_\beta$$ is closed, $$\alpha \in C_\beta$$ for all $$\beta < \alpha,$$ so $$\alpha \in C.$$ To show $$C$$ is unbounded, let $$\alpha < \kappa,$$ and define a sequence $$\xi_i,$$ $$i < \omega$$ as follows: $$\xi_0 = \alpha,$$ and $$\xi_{i+1}$$ is the minimal element of $$\bigcap_{\gamma<\xi_i} C_\gamma$$ such that $$\xi_{i+1} > \xi_i.$$  Such an element exists since by the above, the intersection of $$\xi_i$$ club sets is club. Then $$\xi = \bigcup_{i<\omega} \xi_i > \alpha$$ and $$\xi \in C,$$ since it is in each $$C_i$$ with $$i < \xi.$$