Club filter explained

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 = c ap 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 c up 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 c ap _{\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 = c up _{i<\omega} \xi _{i} > \alpha and \xi \in C, since it is in each C _{i} with i < \xi.

References

• Jech, Thomas, 2003. Set Theory: The Third Millennium Edition, Revised and Expanded. Springer. .