# Principal Filters and Neighborhood Filter

I'm having trouble solving a problem using filters.

The problem:
Given X a topology such that every ultrafilter that converges is principal.

Show that for any collection of open sets $$(U_a)_{a\in I}$$ the intersection $$\underset {a\in I}{\bigcap U_a}$$ is open.

I have tried this approach: Assume the intersection is not open, then it's not empty and there is $$x\in \underset {a\in I}{\bigcap U_a}$$ such that for all neighborhoods $$U$$ of x, $$U\cap X\setminus \underset {a\in I}{\bigcap U_a}\neq\emptyset$$.

Then take $$F_x$$ the neighborhood filter of x, and $$F_x\subseteq F$$ were $$F$$ is an ultrafilter that converges. By the assumption, $$F$$ is a principal filter where $$F_{(y)}=F$$. From here, I am stuck and don't see where to go.

I would like some help and direction. Am I on the right track? My problem now is that I don't see why $$y=x$$ because the space is not Hausdorff.

I suggest the following path to solve the problem. I can elaborate the proof of each step if necessary.

Observation 1. The following statements about a topological space $$X$$ are equivalent:

1. All convergent ultrafilters on $$X$$ are principal.

2. Any convergent filter on $$X$$ has nonempty intersection.

Observation 2. The following statements about a filter $$F$$ and a point $$x \in X$$ are equivalent.

1. There exists a bigger filter $$G \supset F$$ such that $$G$$ converges to $$x$$.

2. $$x \in \bigcap_{A \in F} \overline{A}$$

Corollary. Let $$X$$ satisfy the equivalent conditions from Observation 1 and let $$F$$ be a filter on $$X$$. If $$\bigcap_{A \in F} \overline{A} \ne \emptyset$$, then $$\bigcap_{A \in F} A \ne \emptyset$$.

For the following two observations we fix a collection $$\{U_a\}_{a \in I}$$ of open sets and let $$W = \bigcap_{a} U_a$$.

Observation 3. If $$W$$ is not open, then the set $$B = \{(U_{a_1} \cap \dots \cap U_{a_n}) \setminus W: n \in \mathbb N, a_1, \dots, a_n \in I \}$$ is a basis of a filter. The filter generated by $$B$$ has empty intersection.

Observation 4. If $$x \in W$$ is not an interior point, then $$x \in \overline{(U_{a_1} \cap \dots \cap U_{a_n}) \setminus W}$$ for any finite collection $$a_1, \dots, a_n \in I$$.

The final steps of the proof are trivial: if $$W$$ is not open, then the filter generated by $$B$$ has empty intersection, but nonempty intersection of closures. This contradicts the foregoing corollary, if $$X$$ satisfies the conditions of Observation 1.

• Thanks for the asnwer, manged to proof it by those idea.
– A.P
Commented Aug 1 at 14:41