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 $$x$$.