Point in Closure of Subset of Metric Space iff Limit of Sequence

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 $H$.

Let $a \in A$.

Then $a \in H^-$ iff there exists a sequence $\left\langle{x_n}\right\rangle$ of points of $H$ which converges to the limit $a$.

Proof
From definition of closure, $H^- = H' \cup H^i$.

Suppose that $a \in H^-$.

If $a \in H^i$, then $a \in H$ and so $\left\{{a, a, \ldots}\right\}$ is a sequence in $H$ that converges to $a$.

If $a \in H'$, then by Limit Point is Limit of Convergent Sequence there exist a sequence in $H$ that converges to $a$.

For the converse, let $\left\langle{x_n}\right\rangle$ be a sequence in $H$ that converges to a point $a \in A$.

So let $U$ be an open set that contains $a$.

By the definition of an open set, it follows that there exists an open ball centered at $a$ such that $B_\epsilon \left({a}\right) \subseteq U$.

But then because $\left\langle{x_n}\right\rangle$ converges to $a$ there exist a $m \in \N_{\gt 0}$ such that
 * $d \left({x_m, a}\right) \lt \epsilon \implies H \cap B_\epsilon \left({a}\right) \ne \varnothing \implies H \cap U \ne \varnothing$

Because from Set Intersection Preserves Subsets it is seen that $H \cap B_\epsilon \left({a}\right) \subseteq H \cap U$.

It has been shown that every open set that contains $a$ also contains a point in $H$.

So by Condition for Point being in Closure, it is seen that $a \in H^-$.