Club filter

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 ${\displaystyle i<\xi }$.