# 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 $i<\xi$.

## References

• Jech, Thomas, 2003. Set Theory: The Third Millennium Edition, Revised and Expanded. Springer. ISBN 3-540-44085-2.