Show that $\bar{\bar{A}}=\bar{A}$ Show that $\bar{\bar{A}}=\bar{A}$ by using the following definition of limit point:
$a$ is a limit point of a set $A$ if there exists a sequence $(a_n) \subset A, a_n \neq a$ and $a_n \rightarrow a$. 
I manage to prove $\bar{A} \subset \bar{\bar{A}}$. I stuck at another implication. 
Let $x \in \bar{\bar{A}}$. Then by definition, there exists a sequence $(x_n) \subset \bar{A}, x_n \neq x$ and $x_n \rightarrow x$. My aim is to make sure there are infinitely many $x_n \in A$ so that I can have a subsequence $(x_{n_j})$ in order to conclude $x$ is a limit point of $A$, and hence $x \in \bar{A}$. But I fail to do so. 
Can anyone give some hint?

In your notation, I assume that the space has a metric :
If $a\in \overline{\overline{A}}$ then there exists $
  a_n\neq a \in
  \overline{A}$ s.t. $a_n \rightarrow a\ (n\geq N \Rightarrow d(a_n,a)<\epsilon)$.
Given $\epsilon$ and $a_n$, there exists $b_n\in A$ s.t. $d(a_n,b_n)< \epsilon $. Hence $d(a,b_{n}) \leq  d(a,a_n) + d(a_n,b_n) <2\epsilon $
where $n> N$. For each $\epsilon$ we have $b_n\in A$. So $a$ is a limit point of $A$.  
