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 $$x \in T - H$$.

Then $$T - H$$ 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 with Complement of 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 $$x \in T - H$$ there is an open set $$U_x$$ such that $$x \in U_x$$ and $$U_x \cap H = \varnothing$$.

So $$U_x \subseteq T - H$$.

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