Converging sequence in a metric space

Question

Let $(V,d)$ be a metric space and $A \subseteq V$. I have to proof the following:

$a \in V$ is a limit point of $A$ if and only if there exists a sequence $(x_n)_{n \in \mathrm{N}_0}$ in $A$ with $\lim_{n \to \infty}x_n = a$.

I think I have proven it from left to right in the following way:

Let there exists a sequence $(x_n)_{n \in \mathrm{N}_0}$ in $A$ with $\lim_{n \to \infty}x_n = a$. $x_n \in A$ for all $x_n$ is given, so we only need to prove that there exists a $n$ s.t. $x_n \in B(a; \epsilon)$. The limit of this sequence exists, thus for all $\epsilon >0$ there exists a $N \in \mathrm{N}$ s.t. for $n>N\implies x_n \in B(a; \epsilon)$. Thus $x_n \in B(a; \epsilon) \cap A$ thus $a$ is a limit point of $A$.

Now I have two questions. First of all I want to know whether my proof is correct and secondly I would like to know how to start my proof the other way around. I get confused by how I should define a sequence in $A$ for this part of the proof. I feel that I need to find a way to first define a sequence before I can proof that a row exists that converges to $a$.

Answer 1

We want to prove that: exists $(x_n)_{n \in \mathbb{N}}\subset A$ with $x_n \to a \iff a \in \overline{A}$.

Your proof of the $\implies$ direction is okay. For the other direction, suppose that $a \in \overline{A}$. So for every $n \geq 1$, $B(a, 1/n)\cap A$ is non-empty and you can take an element there - which you'll call $x_n$. Then you can check that $x_n \to a$.

Answer 2

Your proof so far is fine. For the other direction, pick $x_n\in B(a;\frac1n)\cap A$, which must be possible if $a$ is a limit point.