# 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 $\mathbb K$.

$H \subseteq H^-$

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

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

Let $K \in \mathbb K$.

$\ds H^- = \bigcap \mathbb K$

Therefore, from Intersection is Subset: General Result:

$H^- \subseteq K$

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

$\blacksquare$