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

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## 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 $S \setminus F$.

From Set Difference with Subset is Superset of Set Difference:

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

From Subsets of Disjoint Sets are Disjoint:

- $S \setminus F \cap H = \O$

From Set is Open iff Neighborhood of all its Points:

- $S \setminus F$ is an neighborhood of $x$

By definition of the closure of $H$:

- $x \notin H^-$

We have shown that:

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

From Set Difference with Subset is Superset of Set Difference:

- $H^- \subseteq F$

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*... (previous): $\S 4.6$: Closure, Interior, Boundary: Lemma $4.3$