If a point is not isolated then it is a limit point Let $k \in X$, where $X$ is a metric space.  I want to show that, if $k$ is not an isolated point of $U$ where $U \subseteq  X$, then it is a limit point of $u$ if $\exists u_{n} \in U$ whose elements are distinct s.t. $\displaystyle\lim_{n \to \infty}u_{n} = k$.
It seems fairly straightforwards verbally but I was not sure how to prove it. If $k$ is not an isolated point then that means when we consider a neighborhood around it $k$, the neighborhood will be in $U$ in some sort of way. Otherwise if $k$ was an isolated point then $k \cap U = \emptyset$. And then the result would follow. I know my idea was vague. I was just throwing out ideas. Help would be appreciated. 
 A: Let $U$ be a non-empty set in a metric space $(X,d)$, and let $p_o \in U$.
Suppose $p_o$ is not an isolated point of $U$. Then there does not exist $r>0$ so that 
\begin{equation}B(p_o,r) \cap U = \{p_o\} \,.
\end{equation}
In other words, for any $r > 0$ the open ball $B(p_o,r)$ must contain a point $p$ of $U$ which is distinct from $p_o$. Hence $p_o$ is a limit point of $U$ by definition.
Let $n$ be an arbitrary positive integer. Since $1/n>0$, the open ball $B(p_o,1/n)$ contains a point of $U$ distinct from $p_o$ and we denote this point by $p_n$. 
It now follows that for each positive integer $n$, there is a point $p_n \in U$ with $p_n \neq p_o$ and such that $d(p_n, p_o)<1/n$.
Let $\varepsilon>0$ be given. By the Archimedean property, there is a positive integer $N$ so that $N\varepsilon>1$. It now follows that $d(p_n,p_o)<\varepsilon$ whenever $n \geq N$. Thus we have shown that there is a sequence $\{p_n\}_{n \in \mathbb{N}} \subset U$ with $p_n \neq p_o$ for every $n$ and such that $p_n \to p_o$ as $n \to \infty$.
