Closure of Topological Closure equals Closure

Theorem
Let $T$ be a topological space.

Let $H \subseteq T$.

Then:
 * $\left({H^-}\right)^- = H^-$

where $H^-$ denotes the closure of $H$.

Proof
It follows directly from Set is Subset of its Topological Closure that:
 * $H^- \subseteq \left({H^-}\right)^-$

Let $x \in \left({H^-}\right)^-$.

Then from Condition for Point being in Closure, any $U$ which is open in $T$ such that $x \in U$ contains some $y \in H^-$.

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

Hence $x \in H^-$.