Limit point of the set of limit points is in the set of limit points. Let $A'$ be the set of all limit points of $A$
$x ∈ M$ is a limit point (or accumulation point) if
$∀$ $ε > 0$ $∃$ $y$ s.t. $y ∈ B(x,ε)$ ∩ S and $y ≠ x$
Question: Prove $(A')' ⊂ A'$. Give an example such that $(A')'≠ A'$
For the example, Let $A = (1/x,1]$ where $x > 0$
$A' = 0$ and $(A')'$ is the empty set so they are not equal
For the proof, let $x$ be in $(A')'$, then $x$ is a limit point of $A'$
This means there is a point $y$ in $A'$ where $y ≠ x$.
If $y$ is in $A'$, then $y$ is a limit point of $A$.
This means there is a point $z$ in $A$ where $y ≠ z$
I am not sure how to continue the proof after this point.
 A: HINT: $x\in V_y$ where $y\in (A')'$, with $x\in A'$ and $x\neq y$. This is the definition of limit point applied to $A'$, i.e. $y$ is a limit point of $A'$, where $V_y$ is any neighborhood of $y$ (you can consider, by example, any open ball $B(y,\epsilon)$).
Now you must show that for any $x$ exists an $B(x,\epsilon')\subset B(y,\epsilon)$, then $y$ is a limit point of $A$ too, so $(A')'\subset A'$.
A: let $y\in (A^{'})^{'}$ then $\forall r > 0 \  \exists \ x_{r}\neq y $ such that $x_{r}\in (N(y;r)\cap A^{'}) $
$$\Rightarrow x_{r}\in A^{'}$$
$d = d(x_{r},y)$ then 
$0<d<r.$  Choose $\delta =min \{r-d, d\}$ then $N(x_{r};\delta)\cap A \neq \emptyset $, $y\notin N(x_{r};\delta)$ and $N(x_{r}; \delta)\subset N(y;r)$. 
Now $x_{r}\in A^{'}$ $$\Rightarrow \ \  \exists z_{r} \in N(x_{r};\delta) \cap A$$ 
$$\Rightarrow z_{r} \in N(y;r)\cap A \textit{ and }  z_{r}\neq y$$ 
$\Rightarrow y\in A^{'}$
A: $x\in A'$ iff every open set containing  $x$ also contains some $z$ with $x\ne z\in A.$
Suppose $x\in (A')'.$ Let $U$ be an  open set containing $x.$ We show there exists $z$ with $x\ne z \in U\cap A.$
There exists $y\in U\cap A'$ with $y \ne x.$ Since we are in a metric space, there is an open set $V$ with $y\in V$ and $x \not \in V.$  Now $y$ belongs to the open set $U\cap V$, and $y\in A',$ so there exists $z\in (U \cap V)\cap A$....( We can also require $z\ne y$, but that is is not needed.) 
And $z\ne x$ because $z\in V$ and $x \not \in V.$ Therefore $x\ne z\in (U\cap V)\cap A\subset U\cap A.$
