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 $T$.

Let $A$ be a set.

Then $A = H^-$ iff both of the following hold:


 * $(1): \quad A$ is a superset of $H$ that is closed in $T$.
 * $(2): \quad$ If $K$ is a superset of $H$ that is closed in $T$, then $A \subseteq K$.

That is, $H^-$ can be defined as the smallest superset of $H$ that is closed in $T$, in the above sense.

Proof
Define:
 * $\mathbb K := \left\{{K \supseteq H: K}\right.$ is closed in $\left.{T}\right\}$

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

Necessary Conditions
From the sufficient conditions, we have that:


 * From $(1)$, we have that $H^- \in \mathbb K$, so it follows by hypothesis that $A \subseteq H^-$.


 * From $(2)$, it follows that $H^- \subseteq A$.

By Equality of Sets, we have that $A = H^-$, as desired.

Sufficient Conditions
From the definition of closure, we have that $H \subseteq H^-$.

From Closure is Closed, we have that $H^-$ is closed in $T$.

Let $K$ be a superset of $H$ that is closed in $T$.

That is, let $K \in \mathbb K$.

From Set Closure as Intersection of Closed Sets, we have that $\displaystyle H^- = \bigcap \mathbb K$.

Therefore, from Intersection Subset: General Result, it follows directly that $H^- \subseteq K$.