Let us recall that a regular topological space is *semi-stratifiable* if each point $x\in X$ has a countable family of neighborhoods $(U_n(x))_{n\in\omega}$ such that each closed subset $F\subset X$ is equal to the intersection $\bigcap_{n\in\omega}U_n[F]$ of its neighborhoods $U_n[F]=\bigcup_{x\in F}U_n(x)$.

If, moreover, $F=\bigcap_{n\in\omega}\overline{U_n[F]}$, then the space $X$ is called *stratifiable*.

Basic information on stratifiable spaces can be found in this survey paper of Gruenhage.

Definition.A topological space $X$ is called an$\bullet$

$r'$-spaceits set $X'$ of non-isolated point is a retract of $X$;$\bullet$

hereditarily $r'$-spaceif each subspace of $X$ is an $r'$-space.

Observe that each separable $r'$-space has separable set $X'$ of non-isolated points.

It can be shown that each stratifiable space is a (hereditary) $r'$-space.

On the other hand, any uncountable Mrowka-Isbell space $X$ is semi-stratifiable but fails to be an $r'$-space (since it is separable but its set $X'$ of non-isolated points is not separable).

Question.Is each semi-stratifiable hereditary $r'$-space $X$ stratifiable?What will be the answer if we additionally assume that $X$ is hereditarily paracompact?

**Added in Edit.** The hereditary $r'$-space property indeed characterizes the stratifiability at least in the class of scattered spaces of finite scattered height, as shown by the following theorem (that can proved by induction):

Theorem.For a scattered space $X$ of finite scattered height $\hbar(X)$ the following conditions are equivalent:1) $X$ is stratifiable;

2) each non-empty closed subset of $X$ is a $G_\delta$-set in $X$ and is a retract of $X$;

3) $X$ is hereditarily $r'$-space and each closed subset of $X$ is of type $G_\delta$;

4) for every $n<\hbar(X)$ the $n$-th derived set $X^{(n)}$ is a $G_\delta$-set in $X$ and $X^{(n)}$ is a retract of $X$.

The derived sets $X^{(\alpha)}$ of a topological space $X$ are defined for any ordinal $\alpha$ by the recursive formulas:

$\bullet$ $X^{(0)}=X$,

$\bullet$ $X^{(1)}$ is the set of non-isolated points in $X$, and

$\bullet$ $X^{(\alpha)}=\bigcap_{\beta<\alpha}(X^{(\beta)})^{(1)}$ for any ordinal $\alpha>0$.

The *scattered height* $\hbar (X)$ of a topological space $X$ is the smallest ordinal $\alpha$ such that $X^{(\alpha+1)}=X^{(\alpha)}$.

A topological space $X$ is *scattered* if $X^{(\hbar(X))}=\emptyset$.