# Closure of Subset of Metric Space is Intersection of Closed Supersets

## Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Let $H \subseteq A$ be a subset of $A$.

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

Then $H^-$ is the intersection of all closed sets of $M$ of which $H$ is a subset.

## Proof

Let $\mathbb K$ be the set of all closed sets $K$ of $M$ such that $H \subseteq K$.

From Closure of Subset of Closed Set of Metric Space is Subset:

- $\forall K \in \mathbb K: H^- \subseteq K$

From Intersection is Largest Subset:

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

$\Box$

From Closure of Subset of Metric Space is Closed, $H^-$ is closed.

From Subset of Metric Space is Subset of its Closure, $H \subseteq H^-$.

So $H^-$ is, by definition, a closed set of $M$ which contains $H$.

But $\displaystyle \bigcap \mathbb K$ is the intersection of **all** closed sets of $M$ which contain $H$.

So from Intersection is Subset it follows that:

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

$\Box$

Finally, we have that:

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

So by definition of set equality:

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

which is what we needed to prove.

$\blacksquare$

## Sources

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 2.6$: Open Sets and Closed Sets: Exercise $6$