I am reviewing an article on separation axioms (Last part of Theorem 3.7, https://core.ac.uk/download/pdf/82702431.pdf ). I have trouble understanding the following statement:
If $\{x\}'$ is not closed, then its closure is $\overline{\{x\}}$
And from this statement we obtain that $\overline{\{x\}}\cap \{y\}'=\emptyset$.
(The set $\{x\}'$ represents the limit points of $\{x\}$.)
My only hypothesis is that any set of limit points of any two points are separated, that is, if $x,y \in X$ then $\overline{\{x\}'}\cap \{y\}'= \emptyset$ and $\{x\}'\cap\overline{\{y\}'}=\emptyset$.
I have tried to prove that $x\in (\{x\}')'$, since this implies that $\overline{\{x\}} \subseteq \overline{\{x\}'}$ and thus conclude that $\overline{\{x\}}=\overline{\{x\}'}$.
I have also tried to prove that $(\{x\}')'\subseteq \{x\}'$, since this implies that $\overline{\{x\}'}\subset \{x\}' \subset \overline{\{x\}}$, and thus conclude that $\overline{\{x\}}=\overline{\{x\}'}$.
But I'm really confused, I hope you can guide me a little.
Note that $\{x\}' = \overline{\{x\}}\setminus \{x\}$. We have $\{x\}'\subseteq \overline{\{x\}'}\subseteq \overline{\{x\}}$ which implies $\overline{\{x\}'} = \{x\}'$ or $\overline{\{x\}'} = \overline{\{x\}}$, since those are the only sets between $\{x\}' = \overline{\{x\}}\setminus\{x\}$ and $\overline{\{x\}}$.
If $\{x\}'$ is not closed, this of course means that $\overline{\{x\}'} = \overline{\{x\}}$.