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Club filter

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 Ci, and therefore its limit is also in every Ci. To show that it is unbounded, take some β < κ. Let \langle \beta_{1,i}\rangle be an increasing sequence with β1,1 > β and \beta_{1,i}\in C_i for every i < α. Such a sequence can be constructed, since every Ci is unbounded. Since α < κ and κ is regular, the limit of this sequence is less than κ. We call it β2, 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 Ci, and all these sequences have the same limit (the limit of \langle\beta_{j,i}\rangle). This limit is then contained in every Ci, 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, and let C = Δi < κCi. 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 Cβ is closed, \alpha\in C_\beta, so \alpha\in C.


Last updated: 06-01-2005 02:36:18
10-26-2009 08:16:03
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