proof of every filter is contained in an ultrafilter (alternate proof)
Claim 1. Every chain in has an upper bound also in .
Claim 2. Either or (or both) are a filter subbasis.
Now, by Claim 2, if were not an ultrafilter, i.e., if for some subset of we would have neither nor in , then the filter generated or would be finer than , and then would not be maximal.
So is an ultrafilter containing , as intended.
|Title||proof of every filter is contained in an ultrafilter (alternate proof)|
|Date of creation||2013-03-22 17:52:54|
|Last modified on||2013-03-22 17:52:54|
|Last modified by||brunoloff (19748)|