To see that this is a filter, note that since it is obviously both closed and unbounded. If then any subset of containing is also in , since , and therefore anything containing it, contains a club set.
It is a complete filter because the intersection of fewer than 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 appears in every , and therefore its limit is also in every . To show that it is unbounded, take some . Let be an increasing sequence with and for every . Such a sequence can be constructed, since every is unbounded. Since and is regular, the limit of this sequence is less than . We call it , 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 , is an increasing sequence contained in , and all these sequences have the same limit (the limit of ). This limit is then contained in every , and therefore , and is greater than .
To see that is closed under diagonal intersection, let , be a sequence, and let . Since the diagonal intersection contains the intersection, obviously is unbounded. Then suppose and . Then for every , and since each is closed, , so .
|Date of creation||2013-03-22 12:53:11|
|Last modified on||2013-03-22 12:53:11|
|Last modified by||Henry (455)|