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# 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 $\alpha\in C$.

## Mathematics Subject Classification

03E10*no label found*

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