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

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

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

To see that ${\displaystyle \operatorname {club} (\kappa )}$ is closed under diagonal intersection, let ${\displaystyle \langle C_{i}\rangle ,}$ ${\displaystyle i<\kappa }$ be a sequence of club sets, and let ${\displaystyle C=\Delta _{i<\kappa }C_{i}.}$ To show ${\displaystyle C}$ is closed, suppose ${\displaystyle S\subseteq \alpha <\kappa }$ and ${\displaystyle \bigcup S=\alpha .}$ Then for each ${\displaystyle \gamma \in S,}$ ${\displaystyle \gamma \in C_{\beta ))$ for all ${\displaystyle \beta <\gamma .}$ Since each ${\displaystyle C_{\beta ))$ is closed, ${\displaystyle \alpha \in C_{\beta ))$ for all ${\displaystyle \beta <\alpha ,}$ so ${\displaystyle \alpha \in C.}$ To show ${\displaystyle C}$ is unbounded, let ${\displaystyle \alpha <\kappa ,}$ and define a sequence ${\displaystyle \xi _{i},}$ ${\displaystyle i<\omega }$ as follows: ${\displaystyle \xi _{0}=\alpha ,}$ and ${\displaystyle \xi _{i+1))$ is the minimal element of ${\displaystyle \bigcap _{\gamma <\xi _{i))C_{\gamma ))$ such that ${\displaystyle \xi _{i+1}>\xi _{i}.}$ Such an element exists since by the above, the intersection of ${\displaystyle \xi _{i))$ club sets is club. Then ${\displaystyle \xi =\bigcup _{i<\omega }\xi _{i}>\alpha }$ and ${\displaystyle \xi \in C,}$ since it is in each ${\displaystyle C_{i))$ with ${\displaystyle i<\xi .}$