# Superset of Neighborhood in Topological Space is Neighborhood

## Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $x \in S$.

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

Let $N \subseteq N' \subseteq S$.

Then $N'$ is a neighborhood of $x$ in $T$.

That is:

$\forall x \in S: \forall N \in \mathcal N_x: N' \supseteq N \implies N' \in \mathcal N_x$

where $\mathcal N_x$ is the neighborhood filter of $x$.

## Proof

By definition of neighborhood:

$\exists U \in \tau: x \in U \subseteq N \subseteq S$

where $U$ is an open set of $T$.

$U \subseteq N'$

The result follows by definition of neighborhood of $x$.

$\blacksquare$