Jtdylktuy

Let $(X,tau)$ be a topological space and $Asubseteq X$

Show $(overline{A})'=A'$ if $(X,tau)$ for $T_1$.

A $T_1$space is a topological space any $x,yin X$ then there exists an open set $U$ such that $Ucap {x,y}={y}$.

So if $(X,tau)$ is $T_1$, I can find $Uintau$ so that $ain A$ and $ain U$ so there are two options:
1) $Usubset A$ and $Ucap Aneqemptyset$ then $a$ is a limit point. Since $a$ is arbitrary all the points in A are limit points.

2)If $Ucap Ext {A}neqemptyset$ and $Ucap Aneq emptyset$ then there exists $xin U$ so that $xnotin A$ and therefore $x$ is a limit point of $A$.

If only 1) is true then $overline{A}=Aimplies(overline{A})'=A'$

If 2) is true then $Asubset overline{A}$ so that $overline{A}=A'cup A$ so that $Ucap overline{A}=Ucap(Acup A')=(Ucap 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.