In mathematics, particularly in set theory, if \kappa is a regular uncountable cardinal then 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 since it is thus both closed and unbounded (see club set). If then any subset of \kappa containing x is also in 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 is a sequence of club sets where Obviously 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 Let be an increasing sequence with and for every i < \alpha. Such a sequence can be constructed, since every C_i is unbounded. Since and \kappa is regular, the limit of this sequence is less than \kappa. We call it \beta_2, and define a new sequence similar to the previous sequence. We can repeat this process, getting a sequence of sequences where each element of a sequence is greater than every member of the previous sequences. Then for each i < \alpha, is an increasing sequence contained in C_i, and all these sequences have the same limit (the limit of ). This limit is then contained in every C_i, and therefore C, and is greater than \beta. To see that is closed under diagonal intersection, let i < \kappa be a sequence of club sets, and let To show C is closed, suppose and Then for each for all Since each C_\beta is closed, for all so To show C is unbounded, let and define a sequence \xi_i, i < \omega as follows: and \xi_{i+1} is the minimal element of such that Such an element exists since by the above, the intersection of \xi_i club sets is club. Then and \xi \in C, since it is in each C_i with i < \xi.