Derived Sets in arbitrary metric space Edited 
Let us take any set $E$ in a metric space. Is it possible to find a $E'$ and $E''$ such that $E$ and $E'$ are non empty but $E''$ is empty. 
If so, Let us now say $E^n$ is $n$th derived set of $E$. Is a generalization possible when we can find a set $E$ such that $E^n$ is nonempty but $E^{n+1}$ is empty?
P.S. By arbitrary I mean it is at least possible to find one such metric space and one such Set $E$, where the first statement might be satisfied.
 A: In case the metric space has at least one cluster point, you can take as $E$ any sequence that clusters around that point and doesn't cluster anywhere else. The derived set of the sequence will be just the cluster point and obviously the derived set of the one-element set is empty. E.g., take $(1/n)_{n \geq 1} \subset \mathbb R$.
You can iterate this process to higher dimension at least in $\mathbb R^n$. For example, take $E = (1/n, 1/m)_{n,m \geq 1} \subset \mathbb R^2$. Then you have $E' = (1/n, 0)_{n \geq 1} \cup (0, 1/m)_{m \geq 1}$ and $E'' = {(0,0)}$. And so on.
A: $\newcommand{\Cantor}{\mathcal{C}}$A very nice metric space to accomplish this is $\Cantor = \{0,1 \}^{\mathbb{N}}$ consisting of all infinite binary sequences.  (I'll take $\mathbb{N} = \{ 0 , 1 , 2 , \ldots \}$ in the sequel.)  A metric for this space is given as follows: for $\mathbf{x} = ( x_n )_{n \in \mathbb{N}} , \mathbf{y} = ( y_n )_{n \in \mathbb{N}} \in \mathcal{N}$ we define $$d ( \mathbf{x} , \mathbf{y} ) = \begin{cases}
0, &\text{if }\mathbf{x} = \mathbf{y} \\
2^{-(n+1)}, &\text{if }\mathbf{x} \neq \mathbf{y}, n = \min \{ n \in \mathbb{N} : x_n \neq y_n \}.
\end{cases}$$
A basis for this space is obtained by taking all sets of the form $$[s] = \{ \mathbf{x} \in \Cantor : s \sqsubset \mathbf{x} \}$$where $s$ is a finite binary sequence (and $s \sqsubset \mathbf{x}$ means that $s$ is an initial segment of $\mathbf{x}$).
(This space is homeomorphic to the usual Cantor ternary set, and is, unsurprisingly, called the Cantor space.)
A purely formal construction of the required sets is as follows:  For each $n$ let $$E_n = \{ \mathbf{x} \in \Cantor : \mathbf{x} \text{ has at most }n \text{ }1\text{s} \}$$  These sets are easily seen to be closed.  (If $\mathbf{y} \notin E_n$, then $\mathbf{y}$ has at least $n+1$ $1$s, so let $\ell$ be the coordinate of the $(n+1)$st $1$, and let $s$ be the initial segment of $\mathbf{y}$ up to (and including) the $\ell$th coordinate.  Then every $\mathbf{z} \in [s]$ contains at least $n+1$ $1$s, and is not in $E_n$.)
We claim that $( E_{n+1} )^\prime = E_n$.  Note that if $\mathbf{x} \in E_{n+1} \setminus E_n$, then $\mathbf{x}$ has exactly $n+1$ $1$s in it.  Let $\ell$ denote the coordinate of the $(n+1)$st $1$ in $\mathbf{x}$, and let $s$ denote the initial segment of $\mathbf{x}$ up to, and including, the $\ell$th coordinate.  Then $E_{n+1} \cap [s] = \{ \mathbf{x} \}$, and so $\mathbf{x}$ is an isolated point of $E_{n+1}$.  This implies that $(E_{n+1})^\prime \subseteq E_n$.  Conversely, if $\mathbf{x} \in E_n \subseteq E_{n+1}$ then given any initial segment $s$ of $x$ we can find a coordinate $\ell > \mathrm{length} (s)$ such that $x_\ell = 0$.  Define $\mathbf{y}$ so that $$y_n = \begin{cases}
x_n, &\text{if }n \neq \ell \\
1, &\text{if }n = \ell.
\end{cases}$$
Then $\mathbf{y} \in E_{n+1} \cap [s]$ and is distinct from $\mathbf{x}$.  Therefore $\mathbf{x} \in E_{n+1}^\prime$.
(One can go quite a bit further and construct, for each $\alpha < \omega_1$ a closed $E \subseteq \Cantor$ such that $E^{(\alpha)} \neq \emptyset$ but $E^{(\alpha+1)} = \emptyset$.)
The nice thing about $\Cantor$ is that if $X$ is any nonempty perfect Polish space (i.e. a separable complete metric without isolated points) includes a closed homeomorphic copy of $\Cantor$, and so every nonempty perfect Polish space has closed subsets of the kind desired.
