Closure of Subset of Metric Space by Convergent Sequence

Lemma
Let $M$ be a metric space.

Let $C \subseteq M$.

Let $x \in M$.

Let $\map \cl C$ denote the closure of $C$ in $T$.

Then $x \in \map \cl C$ there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$.

Necessary Condition
Suppose there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$.

Let $\epsilon > 0$.

Then by definition:
 * $\exists N \in \N: \forall n > N: x_n \in \map {B_\epsilon} x$

where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.

Since $\forall n: x_n \in C$, it follows that:
 * $\forall \epsilon > 0: \map {B_\epsilon} x \cap C \ne \O$

Hence $x \in \map \cl C$.

Sufficient Condition
Now suppose $x \in \map \cl C$.

By definition of closure:
 * $\forall n \in \N: \exists x_n \in C \cap \map {B_{1 / n} } x$

Thus clearly $\sequence {x_n}$ converges to $x$.