# Closure of Subset of Closed Set of Topological Space is Subset/Proof 2

## Theorem

Let $T$ = $\struct {S, \tau}$ be a topological space.

Let $F$ be a closed set of $T$.

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

Let $H^-$ denote the closure of $H$.

Then $H^- \subseteq F$.

## Proof

Let $x \notin F$.

Then $x$ is in the open set $S \setminus F$.

$S \setminus F \subseteq S \setminus H$
$S \setminus F \cap H = \O$
$S \setminus F$ is an neighborhood of $x$

By definition of the closure of $H$:

$x \notin H^-$

We have shown that:

$S \setminus F \subseteq S \setminus H^-$
$H^- \subseteq F$

$\blacksquare$