Topological Space is Open Neighborhood of Subset

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

Let $H \subseteq S$ be a subset of $S$.

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

Proof
From Topological Space is Open and Closed in Itself, $S$ is open in $T$.

By hypothesis, $H \subseteq S$.

The result follows from Open Superset is Open Neighborhood.