Set is Closed iff Equals Topological Closure/Proof 2

Theorem
Let $T$ be a topological space.

Let $H \subseteq T$.

Then $H$ is closed in $T$ iff $H = \operatorname{cl}\left({H}\right)$.

That is, a closed set equals its closure (which makes semantic sense).

Proof

 * Suppose $H$ is closed in $T$.

Let $x \in T \setminus H$.

Then $T \setminus H$ is an open set by definition, which contains $x$ and no points of $H$.

Thus from Condition for Point being in Closure, $x \notin \operatorname{cl}\left({H}\right)$.

Thus from Intersection of Complement with Subset is Empty it follows that $\operatorname{cl}\left({H}\right) \subseteq H$.

Since $H \subseteq \operatorname{cl}\left({H}\right)$ by definition of closure, $\operatorname{cl}\left({H}\right) = H$.


 * Now suppose $H = \operatorname{cl}\left({H}\right)$.

Then for any $x \in T \setminus H$ there is an open set $U_x$ such that $x \in U_x$ and $U_x \cap H = \varnothing$.

So $U_x \subseteq T \setminus H$.

Thus $\displaystyle T \setminus H = \bigcup_{x \in T \setminus H} U_x$ is open and hence $H$ is closed in $T$.