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Club filter

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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 $i<\xi .$

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Club filter

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