# Regularity and ultrafilter

I read the following result in an article.

Let $$X$$ be a regular space. Let $$\mathcal{M}$$ be free closed ultrafilter on $$X$$. Set $$\mathcal{U=}\left\{ U:U\text{ is open and there exists a }F\in \mathcal{M}\text{ such that }F\subset U\right\}$$. $$\mathcal{U% }$$ is contained in an open ultrafilter $$\mathcal{W}$$. By regularity, $$\bigcap \overline{\mathcal{W}}=\emptyset$$.

Why the regularity implies $$\bigcap \overline{\mathcal{W}}=\emptyset$$?

• Only to clarify terminology: By closed/open ultrafilter you mean a maximal filter in the set of closed/open subsets of $X$? – Dieter Kadelka Sep 1 at 16:08
• By regularity each point will have an open neighborhood whose closure is disjoint to a set in $\mathcal{M}$. The complement of this closure is in $\mathcal{W}$, and the closure of this complement omits the original open set around the point. – Todd Eisworth Sep 1 at 16:32
• @DieterKadelka, yes, I mean a maximal filter in the set of closed/open subsets of $X$. – Mehmet Onat Sep 2 at 7:51

## 1 Answer

Maybe that this answer is essentially the not elaborated answer of Todd Eisworth. Assume that $$x \in \bar{W}$$ for all $$W \in \mathcal{W}$$. Since $$\mathcal{M}$$ is free there is $$M \in \mathcal{M}$$ with $$x \not\in M$$. By regularity of $$X$$ there are open $$U,V \subset X$$ with $$x \in U$$, $$M \subset V$$ and $$U \cap V = \emptyset$$. Therefore $$V \in \mathcal{U}$$, but $$U \not\in \mathcal{U}$$, in particular $$V \in \mathcal{W}$$ and $$x \not\in \bar V$$, since $$U \cap V = \emptyset$$ and $$x \in U$$ open.