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
Let $H^{\complement}$ denote the relative complement of $H$ in $T$.

By definition, we have that $H$ is closed in $T$ iff $H^{\complement}$ is open in $T$.

By Set is Open iff Neighborhood of all its Points, this is equivalent to:
 * $\forall x \in H^{\complement}: \exists U \in \tau: x \in U \subseteq H^{\complement}$

By Empty Intersection iff Subset of Complement, we have that:
 * $U \subseteq H^{\complement} \iff U \cap H = \varnothing$

By Condition for Point being in Closure, it follows that $H^{\complement}$ is open in $T$ iff:
 * $\forall x \in H^{\complement}: x \notin \operatorname{cl}\left({H}\right)$

By the Rule of Transposition, this is equivalent to $\operatorname{cl}\left({H}\right) \subseteq H$.

From Set is Subset of its Topological Closure, we have that $H \subseteq \operatorname{cl}\left({H}\right)$.

By definition of set equality, this is equivalent to:
 * $H = \operatorname{cl}\left({H}\right)$