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.


  • Jech, Thomas, 2003. Set Theory: The Third Millennium Edition, Revised and Expanded. Springer. ISBN 3-540-44085-2.

This article incorporates material from club filter on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.