Subset of Metric Space is Closed iff contains all Zero Distance Points

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

Let $H \subseteq A$.

Then $H$ is closed in $M$ iff:
 * $\forall x \in A: d \left({x, H}\right) = 0 \implies x \in H$

where $d \left({x, H}\right)$ denotes the distance between $x$ and $H$.

Necessary Condition
Let $H$ be closed in $M$.

Let $x \in A: d \left({x, H}\right) = 0 \implies x \in H$.

By Existence of Sequence in Subset of Metric Space whose Limit is Infimum there exists a sequence $\left\langle{a_n}\right\rangle$ of points of $H$ such that:
 * $\displaystyle \lim_{n \mathop \to \infty} d \left({x, a_n}\right) = 0$

So every neighborhood of $x$ contains points of $H$.

If some $a_n \in H$ then $x \in H$.

Otherwise each $a_n$ is different from $x$

Then $x$ is the limit of $\left\langle{a_n}\right\rangle$ and hence of $H$.

Thus $x \in H$ by definition of closed set.

Sufficient Condition
Suppose $H$ is such that:
 * $\forall x \in A: d \left({x, H}\right) = 0 \implies x \in H$

Let $x$ be a limit point of $H$.

Then $d \left({x, H}\right) = 0$.

Thus $H$ contains all its limit points and so is by definition a closed set.