Let $(X,\tau)$ be a topological space and $A\subseteq X$
Show $(\overline{A})'=A'$ if $(X,\tau)$ for $T_1$.
A $T_1$space is a topological space any $x,y\in X$ then there exists an open set $U$ such that $U\cap \{x,y\}={y}$.
So if $(X,\tau)$ is $T_1$, I can find $U\in\tau$ so that $a\in A$ and $a\in U$ so there are two options: 1) $U\subset A$ and $U\cap A\neq\emptyset$ then $a$ is a limit point. Since $a$ is arbitrary all the points in A are limit points.
2)If $U\cap Ext \{A\}\neq\emptyset$ and $U\cap A\neq \emptyset$ then there exists $x\in U$ so that $x\notin A$ and therefore $x$ is a limit point of $A$.
If only 1) is true then $\overline{A}=A\implies(\overline{A})'=A'$
If 2) is true then $A\subset \overline{A}$ so that $\overline{A}=A'\cup A$ so that $U\cap \overline{A}=U\cap(A\cup A')=(U\cap A)\cup (U \cap A)=U \cap A $ so that $(\overline{A})'=A'$
Question:
Is my proof right? If not. Could someone provide me an alternative proof?
Thanks in advance!
Note that it is sufficient to show $A'' \subseteq A'$:
$\overline{A'} = A' \cup A'' \subseteq A'$ in that case, and the reverse inclusion is trivial.
I'll use the $T_1$-ness in that $x \in A'$ iff every open neighbourhood of $x$ contains infinitely many points of $A$. This fact holds in $T_1$ spaces, as is well-known.
If $x \in A''$ then let $O$ be any open neighbourhood of $x$. Then $O$ contains a point $y \in A'$ and then $O$ must thus contain infinitely many points of $A$ (as $O$ is an open neighbourhood of $y$ too and $y \in A'$), and so $x \in A'$ as required.