Club filter

From Wikipedia, the free encyclopedia

In mathematics, particularly in set theory, if $κ$ is a regular uncountable cardinal then $\operatorname{club}(\kappa)$ , the filter of all sets containing a club subset of $κ$ , is a $κ$ -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 $κ$ containing $x$ is also in $\operatorname{club}(\kappa)$ , since $x$ , and therefore anything containing it, contains a club set.

It is a $κ$ -complete filter because the intersection of fewer than $κ$ club sets is a club set. To see this, suppose $\langle C_i\rangle_{i<\alpha}$ is a sequence of club sets where $α < κ$ . 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 $β < κ$ . Let $\langle \beta_{1,i}\rangle$ be an increasing sequence with $β 1,1 > β$ and $\beta_{1,i}\in C_i$ for every $i < α$ . Such a sequence can be constructed, since every $C i$ is unbounded. Since $α < κ$ and $κ$ is regular, the limit of this sequence is less than $κ$ . We call it $β 2$ , 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 $i < α$ , $\langle\beta_{j,i}\rangle$ 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 $β$ .

To see that $\operatorname{club}(\kappa)$ is closed under diagonal intersection, let $\langle C_i\rangle$ , $i < κ$ be a sequence of club sets, and let $C = Δ i < κ C i$ . 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 $β < γ$ . Since each $C β$ is closed, $\alpha\in C_\beta$ for all $β < α$ , so $\alpha\in C$ . To show $C$ is unbounded, let $α < κ$ , and define a sequence $ξ i$ , $i < ω$ as follows: $ξ 0 = α$ , and $ξ i + 1$ is the minimal element of $\bigcap_{\gamma<\xi_i}C_\gamma$ such that $ξ i + 1 > ξ i$ . Such an element exists since by the above, the intersection of $ξ i$ club sets is club. Then $\xi=\bigcup_{i<\omega}\xi_i>\alpha$ and $\xi\in C$ , since it is in each $C i$ with $i < ξ$ .