# Set Closure is Smallest Closed Set

## 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 1

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 $T$.

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:

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

Therefore, from Intersection is Subset: General Result:

- $H^- \subseteq K$

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

$\blacksquare$

## Proof 2

Let $\mathbb K$ be defined as:

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

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:

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

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 is 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 $H$.

$\Box$

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 is Largest Subset: General Result that $V$ is the intersection of all closed sets in $T$ that contain $H$.

That is:

- $\displaystyle V = \bigcap \mathbb K$

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

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

$\Box$

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 $H$.

$\blacksquare$

## Sources

- 1953: Walter Rudin:
*Principles of Mathematical Analysis*... (previous) ... (next): $2.27 c$ - 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 1$: Closures and Interiors