Superset of Neighborhood in Topological Space is Neighborhood

Theorem

Let $T = \struct {S, \tau}$ 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:

$\forall x \in S: \forall N \in \NN_x: N' \supseteq N \implies N' \in \NN_x$

where $\NN_x$ 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 $T$.

By Subset Relation is Transitive:

$U \subseteq N'$

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

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N 3$