Let $A$ be a subset of a topological space $X$. I am interested in establishing under which conditions the following inclusion holds: $A'' \subseteq A'.$
This is certainly false in general: consider the two point space $X=\{x, y\}$ with the indiscrete topology. Then $\{x\}'=\{y\}$ and $\{y\}'=\{x\}$.
On the other hand this is true for $T_1$ spaces. For, if $x \in A''$ any neighborhood $U_x$ of $x$ contains a point $y\in A', y\ne x$; then if $V_y$ is a neighborhood of $y$ which does not contain $x$, $V_y \cap U_x$ is a neighborhood of $y$ and hence contains a point $z \in A, z\ne x, z \in U_x$.
Question is: is $T_0$ axiom enough for the inclusion to hold true? I can't prove this nor find a counterexample.
$A'' \subseteq A'$ happens precisely when derived sets are closed.
This recent answer examimes the question quite thoroughly, and gives a $T_0$ counterexample (Sierpinski's space) as well. It's somewhere between $T_0$ and $T_1$ in nature.