Closure of Subset of Metric Space by Convergent Sequence

Lemma

Let $M$ be a metric space.

Let $C \subseteq M$.

Let $x \in M$.

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

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

Let $\epsilon > 0$.

Then by definition, $\exists N \in \N: \forall n > N: x_n \in N_\epsilon \left({x}\right)$, where $N_\epsilon \left({x}\right)$ is the $\epsilon$-neighborhood of $x$ in $M$.

Since $\forall n: x_n \in C$, it follows that $\forall \epsilon > 0: N_\epsilon \left({x}\right) \cap C \ne \varnothing$.

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

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

By definition of closure, there exists $x_n \in C \cap N_{1 / n} \left({x}\right)$ for every $n \in \N$.

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

$\blacksquare$