Taking the derived set in a topological space twice

Question

For a topological space $(X, \tau)$ and a set $A \subseteq X$, let $\delta A$ denote the derived set of $A$. That is, $\delta A$ is the set of limit points of $A$. In other words, $\delta A$ contains the elements $x \in X$ all of whose open neighbourhoods contain an element $y \neq x$ that is also $A$. In symbols: $$ \delta A = \{ x \in X \mid \forall B \in \tau, \text{ if } x \in B \text{ then } (B \setminus \{ x \}) \cap A \neq \emptyset \}. $$ In [1] it is claimed that $$ \forall A \subseteq X (\delta\delta A \subseteq \delta A) \quad\text{if and only if}\quad \forall x \in X (\delta\delta \{ x \} \subseteq \delta \{ x \}). $$ For the proof, Esakia gives a reference to Kelley's General Topology, but I cannot find a proof there. So I am looking for a reference for this claim, or a proof.

References

[1] L. L. Esakia, Diagonal constructions, the Löb formula and scattered Cantor spaces

Answer 1

First, a fact:

For $C \subseteq X$, $C' \subseteq C$ if and only if $C$ is closed in $X$.

where I am using a prime to denote the set of limit points ("derived set"). The forward implication is straightforward: consider any point $x \notin C$. If $C' \subseteq C$, then $x \notin C'$. Thus, some neighborhood $U$ of $x$ does not intersect $C$; i.e., $U$ is contained in $X - C$. This shows $X - C$ is open in $X$; equivalently, $C$ is closed in $X$. The reverse implication is also straightforward: given $C$ closed in $X$, $X - C$ is open in $X$. Therefore, given $x \notin C$, there is a neighborhood $U$ of $x$ that is contained in $X - C$; i.e., $U$ does not intersect $C$. Therefore, $x \notin C'$. This proves $C' \subseteq C$.

Applying this fact to your question, we instead prove the proposition:

The following are equivalent:

(1) For each $A \subseteq X$, $A'$ is closed.

(2) For each singleton $S = \{ x \}, x \in X$, $S'$ is closed.

(1) implies (2) is trivial. Each singleton set $S$ is obviously a subset of $X$, so by hypothesis, $S'$ is closed.

(2) implies (1) requires a little more work. Let $A \subseteq X$ be arbitrary; we want to show that $A'$ is closed. To do so, we show that $X - A'$ is open. So, let $x \notin A'$ be arbitrary. Since $x$ is not a limit point of $A$, then there is a (open) neighborhood $U$ of $x$ that either (1) does not intersect $A$, or (2) intersects $A$ merely at $x$ itself.

In the first case, $U$ does not intersect $A'$. For if it did, say at $y$, then $U$ being a neighborhood of $y$ and $y \in A'$ together imply that $U$ intersects $A$, contradicting the hypothesis.

In the second case, we have $x \in A$. Letting $S_x = \{ x \}$, we use the fact that the derived set $S_x'$ is closed (from hypothesis), so that $X - S_x'$ is open. Note that $x \notin S_x' \implies x \in X - S_x'$, since any neighborhood of $x$ intersects $S_x$ only at $x$. Therefore, $U \cap (X - S_x')$ is a neighborhood of $x$, and we claim that it does not intersect $A'$. Suppose for the sake of contradiction that $U \cap (X - S_x')$ does intersect $A'$, say at $y$. Since $y \notin S_x'$, there exists some neighborhood $V$ of $y$ that does not intersect $S_x$; particularly, $V$ does not contain $x$. The neighborhood $U \cap V$ of $y$ does not intersect $A$, since $(U \cap V) \cap A = (U \cap A) \cap V = \{ x \} \cap V = \emptyset$. Therefore, $y \notin A'$, contrary to hypothesis.

Either way, there exists a neighborhood of $x \notin A'$ that is disjoint from $A'$, which proves that $A'$ is closed, as desired.