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Stream: deprecated: logic

Topic: Filters

সায়ন্তন রায় (Jun 21 2020 at 13:34):

I was recently going through the notion of filters on a set. The definition of an ultrafilter in the book says that an ultrafilter is a maximal proper filter. Then it is claimed that if $F$ is a proper filter then $F$ is equal to the intersection of all the ultrafilters that contain $F$. How do I show that the intersection of all the ultrafilters that contain $F$ is contained in $F$?

Lê Thành Dũng (Tito) Nguyễn (Jun 21 2020 at 14:01):

Suppose that $F$ is a proper filter on $\mathcal{P}(X)$ and let $S \subset X$ be a non-empty set that is not in $F$. Consider $F ' = \{S'' \mid \exists S' \in F : (S' \setminus S) \subset S''\}$. This $F'$ is a proper filter, containing the complement of $S$, and it can be extended to some ultrafilter (by https://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem#The_ultrafilter_lemma ). So there is an ultrafilter that contains $F$ and not $S$.