# 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 $\operatorname{cl} \left({C}\right)$ denote the closure of $C$ in $T$.

Then $x \in \operatorname{cl} \left({C}\right)$ iff there exists a sequence $\left \langle {x_n} \right \rangle$ in $C$ which converges to $x$.

## Proof

### Necessary Condition

Suppose there exists a sequence $\left \langle {x_n} \right \rangle$ in $C$ which converges to $x$.

Let $\epsilon > 0$.

Then by definition:

$\exists N \in \N: \forall n > N: x_n \in B_\epsilon \left({x}\right)$

where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$ in $M$.

Since $\forall n: x_n \in C$, it follows that:

$\forall \epsilon > 0: B_\epsilon \left({x}\right) \cap C \ne \varnothing$

Hence $x \in \operatorname{cl} \left({C}\right)$.

$\Box$

### Sufficient Condition

Now suppose $x \in \operatorname{cl} \left({C}\right)$.

By definition of closure:

$\forall n \in \N: \exists x_n \in C \cap B_{1 / n} \left({x}\right)$

Thus clearly $\left \langle {x_n} \right \rangle$ converges to $x$.

$\blacksquare$