# Neighborhood filter is ultrafilter?

Is the nbhood filter of a point $x$ an ultrafilter?

It seems to me that this is true in metric space: If $F_x$ is a nbhood filter no sets can be added to it. If a set $Y$ has non-empty intersection with all nbhoods of $x$ then $x$ is in the closure of the set $Y$. If $Y$ is added the resulting collection is not a filter because all intersections are added and there exists an intersection which is not a nbhood. For example in $\mathbb R$ $x=0$ and $Y=[0,1]$.

In which topological spaces is the nbhood filter an ultrafilter?

## 2 Answers

Let $X$ be a space, $x\in X$, and $\mathscr{N}(x)$ the nbhd filter at $x$. Clearly $x\in\bigcap\mathscr{N}(x)$, so $\mathscr{N}(x)$ is a fixed filter. Thus, $\mathscr{N}(x)$ is an ultrafilter if and only if $\mathscr{N}(x)=\{A\subseteq X:x\in A\}$, the principal ultrafilter over $x$. This is the case iff $\{x\}\in\mathscr{N}(x)$, i.e., iff $x$ is an isolated point of $X$.

Here I’m taking neighborhood in its wider sense: $N$ is a nbhd of $x$ if $x\in\operatorname{int}N$. If you mean the filter of open nbhds of $X$, that’s an ultrafilter iff $X$ bears the discrete topology.

Let $X$ be a topological space and $x \in X$. Suppose that the neighborhood filter $\mathfrak{F}$ at $x$ is an ultrafilter. Let $U \in \mathfrak{F}$. Then $U= V \cup \{x\}$ with $V=U \backslash \{x\}$, so $V \in \mathfrak{F}$ or $\{x\} \in \mathfrak{F}$; but $x \notin V$ so $V \notin \mathfrak{F}$, thus $\{x\} \in \mathfrak{F}$. You deduce that $\mathfrak{F}$ is principal: $\mathfrak{F}= \{ W \subset X : x \in W\}$. Consequently, $x$ is an isolated point.

• I do not understand the last sentence: if $X = [0,1] \cup \{2\}$ then the nbhood filter at $2$ is principal but the topology on $X$ is not discrete. – dolan Feb 24 '13 at 13:12
• @user58975: I wanted to say that if any neighborhood filter is an ultrafilter, then $X$ is discrete. But it's not clear in my answer, in fact $x$ is just an isolated point. – Seirios Feb 24 '13 at 18:19