Open Superset is Open Neighborhood

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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.

$\blacksquare$