Set Closure is Smallest Closed Set/Topology

Theorem
Let $T$ be a topological space.

Let $H \subseteq T$.

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

Then $H^-$ can alternatively be defined as the smallest closed set that contains $H$, in the sense that:

If $K$ is a closed set such that contains $H \subseteq K$, then $H^- \subseteq K$.

Proof
Let $\mathbb K$ be defined as:
 * $\mathbb K := \left\{{K \subseteq T: H \subseteq K, K \text{ closed}}\right\}$

That is, let $\mathbb K$ be the set of all closed sets of $T$ which contain $H$.

Then from Set Closure as Intersection of Closed Sets, we have:
 * $\displaystyle H^- = \bigcap_{K \in \mathbb K} K$

That is, $H^-$ is the intersection of all the closed sets of $T$ which contain $H$.


 * Let $K \subseteq T$ such that $K$ is closed in $T$ and $H \subseteq K$.

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

Then from Intersection Subset it follows directly that $H^- \subseteq K$.

So $H^-$ is a subset of any closed set in $T$ which contains $H$, and so is the smallest closed set that contains $H$.

Let $V$ be the smallest closed set that contains $H$.

Let $K$ be closed in $T$ such that $H \subseteq K$.

Then by definition $V \subseteq K$.

It follows from Intersection Largest: General Result that $V$ is the intersection of all closed sets in $T$ that contain $H$.

That is:


 * $\displaystyle V = \bigcap_{K \in \mathbb K} K$

where $\mathbb K := \left\{{K \subseteq T: H \subseteq K, K \text{ closed}}\right\}$.

From Set Closure as Intersection of Closed Sets it follows that $V = H^-$.

The implication has been demonstrated to hold in both directions, so the closure of $H$ in $T$ can be defined as the smallest closed set that contains $H$.