Closure of Subset of Closed Set of Topological Space is Subset

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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 1

\(\displaystyle H\) \(\subseteq\) \(\displaystyle F\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle H^-\) \(\subseteq\) \(\displaystyle F^-\) Topological Closure of Subset is Subset of Topological Closure
\(\displaystyle \) \(=\) \(\displaystyle F\) Set is Closed iff Equals Topological Closure

$\blacksquare$

Proof 2

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:

$H^- \subseteq F$

$\blacksquare$

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