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?
 A: 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 $A$.  
A: Hint: $\overline {\overline A } $ is the smallest close set containing $\overline{A}$. But $\overline{A}$ is also a close set and $\overline{A}\subset\overline{A}$. So, $\overline {\overline A } \subset \overline{A}$.
A: Since this question has the general topology tag, I think we should have an example of a topological space in which the result does not hold.
Let $X = (\mathbb{N} \times \mathbb{N}) \cup (\mathbb{N} \times \{*\}) \cup \{(*,*)\}\!$. A subset $U$ of $X$ is open if the following two statements are true.


*

*For every $m \in \mathbb{N}$, if $(m,*) \in U$ then $(m,n) \in U$ for all but finitely many $n \in \mathbb{N}$.

*If $(*,*) \in U$ then $(m,*) \in U$ for all but finitely many $m \in \mathbb{N}$.


I leave it as an exercise to check that these open sets form a topology on $X$.
Let $A = \mathbb{N} \times \mathbb{N}$. It follows immediately from the definition of the open sets that $\lim_{n\to\infty} (m,n) = (m,*)$ and $\lim_{m\to\infty} (m,*) = (*,*)$ and so $(*,*) \in \bar{\bar A}$.
We claim that $(*,*) \notin \bar A$. Suppose $((a_n,b_n))_{n \in \mathbb{N}}$ is a sequence in $A$. If there exists $m \in \mathbb{N}$ such that $a_n = m$ for infinitely many $n$ then $X \setminus \{(m,x) \mid x \in \mathbb{N} \cup \{*\}\}$ is an open neighbourhood of $(*,*)$ that doesn't contain any tail of the sequence. Otherwise for every $m \in \mathbb{N}$ there are only finitely many $n$ such that $a_n = m$ and so $X \setminus \{(a_n,b_n) \mid n \in \mathbb{N}\}$ in an open neighbourhood of $(*,*)$ that doesn't contain any tail of the sequence.
