proof of alternative characterization of ultrafilter
Proof that implies or
Once we show that implies , this result will follow immediately.
On the one hand, suppose that and that there exists a such that is empty. Then . Since is a filter and , this implies that .
On the other hand, suppose that and that is not empty for any in . Then would be a filter subbasis. The filter which it would generate would be finer than . The fact that is an ultrafilter means that there exists no filter finer than . This contradiction shows that, if , then there exists a such that is empty. But this would imply that which, in turn would imply that .
Proof that is an ultrafilter.
Assume that is a filter, but not an ultrafilter and that implies or . Since is not an ultrafilter, there must exist filter which is strictly finer. Hence there must exist such that . Set . Since and , it follows that . Since , it is also the case that . But as well; since is a filter, . This is impossible because is empty. Therefore, no such filter can exist and must be an ultrafilter.
Proof of generalization to
On the one hand, since implies , the condition will also imply that is an ultrafilter.
On the other hand, if , there must exists and such that . If is assumed to be a filter, implies that . Likewise, implies that . Hence, if is a filter such that implies that either or , then is an ultrafilter.
Proof of first proposition regarding finite unions
Let and let . For each between and , we have . Hence, either or for each between and . Next, consider three possibilities:
: Since , it follows that .
: Since , it follows that . Because , it follows that .
and : There must exist an such that and . Since , . Since is a filter, . But also which implies that .
This examination of cases shows that if , then there must exist an such that . It is also easy to see that this is unique — If and and , then , but this cannot be the case since is a filter.
Proof of second proposition regarding finite unions
There exist sets such that and . By the result just proven, there exists an such that . Since is a filter, implies . Note that we can no longer assert that is unique because the ’s no longer are required to be pairwise disjoint.
|Title||proof of alternative characterization of ultrafilter|
|Date of creation||2013-03-22 14:42:23|
|Last modified on||2013-03-22 14:42:23|
|Last modified by||rspuzio (6075)|