Closure of Topological Closure equals Closure

Theorem
Let $T$ be a topological space.

Let $H \subseteq T$.

Then $\operatorname{cl}\left({\operatorname{cl}\left({H}\right)}\right) = \operatorname{cl}\left({H}\right)$.

Proof
From the definition of closure, $\operatorname{cl}\left({H}\right)$ is the union of $H$ and its limit points.

Hence $\operatorname{cl}\left({\operatorname{cl}\left({H}\right)}\right)$ is the union of $\operatorname{cl}\left({H}\right)$ and its limit points.

It follows directly from Subset of Union that $\operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({\operatorname{cl}\left({H}\right)}\right)$.

Let $x \in \operatorname{cl}\left({\operatorname{cl}\left({H}\right)}\right)$.

Then from Condition for Point being in Closure, any $U$ which is open in $T$ such that $x \in U$ contains some $y \in \operatorname{cl}\left({H}\right)$.

If we consider $U$ as an open set containing $y$, it follows that $U \cap H \ne \varnothing$.

Hence $x \in \operatorname{cl}\left({H}\right)$.