number of ultrafilters
If is finite then each ultrafilter on is principal, and so there are exactly ultrafilters. In the rest of the proof we will assume that is infinite.
Let be the set of all finite subsets of , and let be the set of all finite subsets of .
For each define , and . For each define .
Let , and suppose and , so that we have . For and let be such that either or . This is always possible, since . Let , and put . Then , for . If for some we have , then for some , which is impossible, as is in one of these sets but not the other. So , and therefore . So . This shows that any finite subset of has nonempty intersection, and therefore can be extended to an ultrafilter .
Suppose are distinct. Then, relabelling if necessary, is nonempty. Let . Then , but since . So and are distinct for distinct and . Therefore is a set of ultrafilters on . But , so has the same number of ultrafilters as . So there are at least ultrafilters on , and there cannot be more than as each ultrafilter is an element of . ∎
The number of topologies on an infinite set is .
Let be an infinite set. By the theorem, there are ultrafilters on . If is an ultrafilter on , then is a topology on . So there are at least topologies on , and there cannot be more than as each topology is an element of . ∎
|Title||number of ultrafilters|
|Date of creation||2013-03-22 15:51:49|
|Last modified on||2013-03-22 15:51:49|
|Last modified by||yark (2760)|