club filter

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 obviously both closed and unbounded. 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, and let $C=\Delta_{i<\kappa}C_{i}$ . Since the diagonal intersection contains the intersection, obviously $C$ is unbounded. Then suppose $S\subseteq C$ and $\sup(S\cap\alpha)=\alpha$ . Then $S\subseteq C_{\beta}$ for every $\beta\geq\alpha$ , and since each $C_{\beta}$ is closed, $\alpha\in C_{\beta}$ , so $\alpha\in C$ .

Title	club filter
Canonical name	ClubFilter
Date of creation	2013-03-22 12:53:11
Last modified on	2013-03-22 12:53:11
Owner	Henry (455)
Last modified by	Henry (455)
Numerical id	5
Author	Henry (455)
Entry type	Definition
Classification	msc 03E10
Defines	club filter