If all derived sets are closed, is the space T$_1$?

Question

In a topological space the set of all limit points of any subset is always closed. Is it necessarily a T$_1$ space? At first I proved that if a space is T$_1$ then the set of all limit points of any subset is closed. Then I thought about the converse. I worked on it for some time but could neither prove it nor found any counterexample.

Answer 1

Counterexamples: Let $X$ be a totally ordered set, topologized so that, for every $U\subseteq X,$ $$U\text{ is open}\iff\forall u\forall x\ (x\gt u\in U\implies x\in U).$$ You can easily verify that $X$ has the property "all derived sets are closed," and that $X$ is not $\text{T}_1$ if it has two or more points. (If $X$ has exactly two points, it's the so-called Sierpiński space.)

A topological space is $\text{T}_1$ iff every one-point subset is closed. A subset of a topological space is locally closed iff it's the intersection of an open set and a closed set. In the space $X$ described above, every one-point set is locally closed; this is a weakening of the $\text{T}_1$ property.

Theorem. For any topological space $X$ the following statements are equivalent:
(1) for every set $A\subseteq X$ the derived set $A'$ is closed;
(2) every one-point subset of $X$ is locally closed.

Proof.
$\underline{(1)\implies(2)}:$ For any point $a\in X$ we have $\{a\}=\overline{\{a\}}\setminus\{a\}',$ where $\overline{\{a\}}$ and $\{a\}'$ are closed sets, so $\{a\}$ is locally closed.

$\underline{(2)\implies(1)}:$ Let a set $A\subseteq X$ be given; we will show that $A'$ is closed by showing that any point $x\notin A'$ has an neighborhood which is disjoint from $A'.$

Consider any point $x\notin A'.$ Since $\{x\}$ is locally closed, we can find an open set $U$ and a closed set $F$ such that $\{x\}=U\cap F.$ Since $x\notin A',$ we can find an open neighborhood $V$ of $x$ such that $V\cap(A\setminus\{x\})=\emptyset.$ Then $U\cap V$ is an open neighborhood of $x$ which is disjoint from $A',$ because $$(U\cap V)\cap F\subseteq U\cap F=\{x\}\text{ which is disjoint from }A',$$ and $$(U\cap V)\setminus F\subseteq V\setminus F\subseteq V\setminus\{x\}\text{ which is disjoint from }A'.$$