Closure of Subset of Metric Space is Closed

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 a closed set of $M$.

Proof
Let $\overline {\paren{H^-}}$ denote the complement of $H^-$.

Let $x \in \overline {\paren{H^-}}$.

By definition of the closure of $H$:
 * $x$ is not a limit point of $H$.

So:
 * $\exists \epsilon \in \R_{> 0} : \paren{\map {B_\epsilon} x \setminus \set x} \cap H = \O$

From Intersection with Set Difference is Set Difference with Intersection:
 * $\paren{\map {B_\epsilon} x \cap H} \setminus \set x = \O$

From Set Difference with Superset is Empty Set:
 * $\map {B_\epsilon} x \cap H \subseteq \set x$

By definition of the closure of $H$:
 * $x$ is not an isolated point of $H$.

So:
 * $\map {B_\epsilon} x \cap H \ne \set x$

Thus:
 * $\map {B_\epsilon} x \cap H = \O$

From Empty Intersection iff Subset of Complement:
 * $\map {B_\epsilon} x \subseteq \overline {\paren{H^-}}$

It follows that $\overline {\paren{H^-}}$ is open in $M$.

Thus $H^- $ is closed in $M$ by definition.