Set of limit points is closed in a general topological space Let $(X, \mathcal{T})$ be a topological space. I came across a question asking for the proof of the fact that the set of limit points $S'$ of any subset $S \subseteq X$ is closed assuming the space is Hausdorff. However, is this always true for a general topological space that is not necessarily Hausdorff?
I am expecting the answer is no. To find a counter example, I tried to use the topology $\mathcal{T} = \{\emptyset, \{1\},\{1,2\},\{1,2,3\}\}$ on $X = \{1,2,3\}$, however I have not been able to find an example of a set whose set of limits points is not closed in this case.
I would appreciate a hint to find a counter example/a proof that none exists rather than a full solution. Thank you very much.
 A: If $X$ is a $T_1$ space then $S'$ is closed. 
Suppose by contradiction that  $p\in \overline {S'}$ \ $S'.$ Since $S'\subset \bar S$ we have $p\in \bar S.$
Now $p\not \in S'$ implies there is an open $U$ with $p\in U$ and $U\cap S\subset \{p\}.$
But $U$ must contain $q\in S'$ because $p\in \overline {S'}$, and by hypothesis, $p\ne q.$  Then there is an open $V$ with $q\in V$, and $p \not \in V$, because $X$ is $T_1.$  The open set $U\cap V$ contains $q\in S'$, so there exists $r\in U\cap V\cap S$ with $r\ne q.$ BUT then $p\ne r\in S\cap U,$ contrary to $U\cap S \subset \{p\}$.
As Opn Ball has already pointed out, the property does not  hold for the coarse topology on $X$ if $X$ has at least 2 points: For $S=\{p\}\subset X$ we have $S'=X$ \ $\{p\}$ and $\overline {S'}=X\ne S'.$
A: Try any set with cardinality at least $2$, equipped with the indiscrete topology.
In detail, let $X=\{a,b\}$ with topology $\tau=\{\emptyset,X\}$. If we take $A=\{a\}$, then $b$ is a limit point of $A$.  But $a$ is not a limit point of $A$ since $a$ is an isolated point in $A$. (A nbhd of $a$ cannot contain a point of $A$ different from $a$, as there are no other points.)  This shows that the derived set is $A'=\{b\}$, which is not closed in $X$.
A: If $X$ is not $T_1$, it might not work as the following example shows. $X=\{a,b,c\}$ and $\tau=\{\emptyset, \{a,b\},X\}$. If we take $A=\{a\}$ then $A'=\{b,c\}$ which is not closed since its complement $\{a\}$ is not in $\tau$. 
