complete ultrafilter and partitions
If is an ultrafilter on a set , then
We prove the case of -completeness; the case of arbitrary infinite cardinality follows closely. For the direction, let be a partition of into many pieces, all of which do not belong to , and write to illustrate this partition. Now, . Since, by our assumption, each of the do not belong to , we have for each as is an ultrafilter. Thus, by -completeness. This, however, means , contradicting the definition of a filter.
for each , put if is the least index for which .
It is easy to verify that each belongs to a unique , the collection of ’s is indeed a partition of .
Along with , partitions into many pieces. A (unique) piece of this partition belongs in : or . But, by the definition of . This excludes the possibility for the former to belong in (cf. alternative characterization of filter) and so .
Thus, starting from an arbitrary collection of -many members of , we have identified a partition of for which the unique piece which belongs to is . Therefore, is -complete.
|Title||complete ultrafilter and partitions|
|Date of creation||2013-03-22 18:55:52|
|Last modified on||2013-03-22 18:55:52|
|Last modified by||yesitis (13730)|