Prove that $(\overline A)'=A'$ in T1 space Let $(X,\tau)$ a $T1$ topological space, and $A\subset X$. We denote $A'=\{x\in X : x \text{ is a limit point of } A\}$
I have to prove the following equality: $(\overline A)'=A'$.
Proving that $A'\subset (\overline{A})'$ is trivial but I need help to  start proving $(\overline A)'\subset A'$ . 

Indeed $A \subseteq \overline{A} \implies A' \subseteq \overline{A}'$. 
Now let $x \in \overline{A}'$. Let's check that $x \in A'$. Let $U$ be an open set, $U\ni x$. We have that $(U \setminus \{x\}) \cap \overline{A} \neq \varnothing$, so we can take $y$ in that intersection. Now $U \setminus\{x\}$ is an open set containing $y$, because the space is $T_1$, and $y \in \overline{A}$, so $(U \setminus\{x\})\cap A \neq \varnothing$, and $x \in A'$, as wanted.
