# Neighborhood of Point in Metrizable Space contains Closed Neighborhood

## Theorem

Let $T = \struct {S, \tau}$ be a metrizable topological space.

Let $x \in S$ be an arbitrary point of $T$.

Let $N$ be a neighborhood of $x$.

Then $N$ has as a subset a neighborhood $V$ of $x$ such that $V$ is closed.

## Proof

Since $N$ is a neighborhood of $x$, there exists an open set $U \subseteq N$ containing $x$ by definition.

As $\struct {S, \tau}$ is metrizable, the set $U$ is open with respect to some metric space $\struct {S, d}$.

Hence:

$\exists \epsilon > 0: \map {\mathcal B_\epsilon} x \subseteq U$

where $\map {\mathcal B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.

Taking $\zeta = \dfrac \epsilon 2$, we have that the closed $\zeta$-ball of $x$ is contained in $\map {\mathcal B_\epsilon} x$, and hence is a subset of $U \subseteq N$.

This closed ball is a closed neighborhood of $x$ in $\struct {S, \tau}$, and is a subset of $N$.

Hence the result.

$\blacksquare$