# Open Superset is Open Neighborhood

## Theorem

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

Let $U \in \tau$ be an open set of $T$.

Let $H \subseteq U$.

Then $U$ is an open neighborhood of $H$.

## Proof

By hypothesis, $U$ is open in $T$.

Also by hypothesis, $H \subseteq U$.

The result follows by definition of open neighborhood.

$\blacksquare$