Open Superset is Open Neighborhood

From ProofWiki
Jump to navigation Jump to search


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 $H$.


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

Also by hypothesis, $H \subseteq U$.

The result follows by definition of open neighborhood.