Prove every derived set is closed if this is the case for singleton sets Suppose that for each $x \in X$, the set of accumulation points of $\{ x \}$ is closed. Then for each $S \subseteq X$, the set of its accumulation points is closed.
This is the last part of exercise $D$ of chapter 1. I managed to solve all the previous points (not indicated here), but unfortunately this one remains open for me. Thank you in advance!
 A: Suppose $S \subseteq X$ and $x \notin S'$. We need to find some open $U$ containing $x$ and such that $U$ misses $S'$. 
$x \notin S'$ means that there exists some open $O$ containing $x$ such that $O \cap S \subseteq \{x\}$.
If $O \cap S = \emptyset$, we can pick $U = O$.
If $O \cap S = \{x\}$, then use that $X \setminus \{x\}'$ is open...
A: I wonder if the following proof has any issue.
Notation:


*

*$A^d: \mbox{derived set of } A$

*$A^c: \mbox{closure of } A$


Lemma: $A \mbox{ is closed} \Leftrightarrow A^d \subset A$.
Proof of the Lemma:


*

*by definition of "closed": $A = A^c$

*by definition of "closure": $A^c = A\cup A^d$.


Therefore, $A$ is closed $\Leftrightarrow A = A\cup A^d \Leftrightarrow A^d\subset A$. $\blacksquare$
By the lemma, to show every $A^d$ is closed, it suffices to show that $A^{dd} \subset A^d$.
Proof:
By a property of derived set, $(A\cup B)^d = A^d \cup B^d$, we have


*

*$A^{dd} = \bigl(\bigcup_{x\in A} \{x\}\bigr)^{dd} = \bigcup_{x\in A} \{x\}^{dd}$

*$A^{d} = \bigl(\bigcup_{x\in A} \{x\}\bigr)^{d} = \bigcup_{x\in A} \{x\}^{d}$


Since each $\{x\}^d$ is closed, we have $\{x\}^{dd} \subset \{x\}^d \Rightarrow A^{dd} \subset A^d. \blacksquare$
