Intersection of the closures of the neighborhoods of a set is the closure of the set

Question

Let $X$ be a topological space. Let $\{x\} \subset X$ and $N(x)$ be the neighborhood filter of $x$. Is $\cap_{n \in N(x)} \bar n = \overline {\{x\}}$?

It at least that $\overline {\{x\}} \subset \cap_{n \in N(x)} \bar n $, for if $n \in N(x), \{x\} \subseteq n$ thus $n \subseteq \bar n$ and $\{x\} \subseteq \overline {\{x\}}$ and thus $\{x\} \subseteq \bar n \cap \overline {\{x\}}$. This is closed, and smaller or equal to $\overline {\{x\}}$, thus $\overline {\{x\}} \subset \cap_{n \in N(x)} \bar n$.

Answer 1

$\newcommand{\cl}{\operatorname{cl}}$It’s false in general. Let $X$ be an infinite space with the cofinite topology. Then $\cl U=X$ for each non-empty $U\in\tau$, so $$\bigcap_{N\in\mathscr{N}(x)}\cl N=X\ne\{x\}=\cl\{x\}$$ for each $x\in X$.

It’s true if $X$ is $T_2$. In that case let $x\in X$, and let $y$ be any other point of $X$. There are disjoint open sets $U$ and $V$ such that $x\in U$ and $y\in V$. Clearly $V\cap\cl U=\varnothing$, so $$y\notin\cl U\supseteq\bigcap_{N\in\mathscr{N}(x)}N\;,$$ and since $y\in X\setminus\{x\}$ was arbitrary, $$\bigcap_{N\in\mathscr{N}(x)}N=\{x\}\;.$$

Answer 2

Not necessarily. Let X be a set containing the distinct points $a$ and $b$. $(X,\{{\emptyset},\{b\},X\})$ is a topological space for which the equality does not hold.

The closure of $\{a\}$ is $X - \{b\}$ and $X$ is the only neighborhood of $a$.

But I think in all uniform spaces the equality holds.