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