does $(\overline{E})^{'}= E^{'} \cup ( E^{'})^{'}$ holds? My question is as follows:
Suppose $E$ is a set in metric space $X$, let $\overline{E}$ denote the closure of E, let $E^{'}$ be the set of all the limit points of $E$. We all know that  $\overline{E}=E\cup E^{'} $ Then my question is: Does the following equality hold?

$(\overline{E})^{'}= E^{'} \cup ( E^{'})^{'}$

if not, can you give me an exception in which the equality does not hold? 
Thanks so much! 

We have more. In a Hausdorff space (even in a $T_1$ space, if you already know what that is), $x$ is a limit point of $A$ if and only if every neighbourhood of $x$ contains infinitely many points of $A$. Thus in such spaces, we have
$$(\overline{E})' = E'.$$
Since evidently $(E')' \subset (\overline{E})'$, the equality holds in Hausdorff (or $T_1$, nore generally) spaces, in particular in metric spaces.
