Open Superset is Open Neighborhood
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Theorem
Let $T = \struct {S, \tau}$ 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.
$\blacksquare$