Subset of Metric Space contains Limits of Sequences iff Closed

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $H \subseteq A$.

Then $H$ is closed in $M$ :
 * for each sequence $\sequence {a_n}$ of points of $H$ that converges to a point $a \in A$, it follows that $a \in H$.

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

Suppose that:
 * $\ds \lim_{n \mathop \to \infty} a_n = a$

and:
 * $\forall n \in \N_{>0}: a_n \in H$

If the set $\set {a_1, a_2, \ldots}$ is infinite then every neighborhood of $a$ contains infinitely many points of $H$.

Thus $a$ is a limit point of $H$.

So by definition of closed set, $a \in H$.

On the other hand, if $\set {a_1, a_2, \ldots}$ is finite, then for some $N \in \N$:
 * $n, m > N \implies a_n = a_m$

Since:
 * $\ds \lim_{n \mathop \to \infty} a_n = a$

Then:
 * $\forall n > N: \map d {a_n, a} = 0$

Thus:
 * $a_n = a$

and so:
 * $a \in H$

Sufficient Condition
Let $H$ be a set such that:
 * for each sequence $\sequence {a_n}$ such that $\ds \lim_{n \mathop \to \infty} a_n = a$, it follows that $a \in H$.

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

By, $b$ is the limit of a convergent sequence of points of $H$.

By hypothesis, $b \in H$.

Thus $H$ is a closed set by definition.