Set Closure is Smallest Closed Set/Topology

Theorem
Let $T$ be a topological space.

Let $H \subseteq T$.

Let $H^-$ denote the closure of $H$ in $T$.

Then $H^-$ is the smallest superset of $H$ that is closed in $T$.

Proof
Define:
 * $\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$

That is, let $\mathbb K$ be the set of all supersets of $H$ that are closed in $T$.

The claim is that $H^-$ is the smallest set of $\mathbb K$.

From Set is Subset of its Topological Closure:
 * $H \subseteq H^-$

From Topological Closure is Closed, $H^-$ is closed in $T$.

Thus $H^- \in \mathbb K$.

Let $K \in \mathbb K$.

From Set Closure as Intersection of Closed Sets:
 * $\ds H^- = \bigcap \mathbb K$

Therefore, from Intersection is Subset: General Result:
 * $H^- \subseteq K$

Thus by definition $H^-$ is the smallest set of $\mathbb K$.