Prove : $x$ is an isolated point of $F$ iff $F − \{x\}$ is still closed. I have a question in the following statement:
Let $F$ be closed and $x\in F$. Then $x $ is an isolated point of $F$
if and only if $F − \{x\}$ is still closed.
To show that it's closed, I've shown the missing inclusion, as another inclusion is obvious. But, considering that the set is closed, I can't see which isolated point.

A set is closed iff it contains all of it's limit points. i.e $\bar{A}=A$ .
Since $x$ is isolated . There exist a nbd $U$ such that $U\cap F\setminus\{x\}=\phi$.
Hence $x$ is not a limit point of $F$.
So if $F$ is closed then $x\notin F'$ and $F'\subset F$. Hence $F'=F'\setminus\{x\} = (F\setminus\{x\})'\subset F\setminus\{x\}$ . So $F\setminus\{x\}$ is closed. $A'$ denotes derived set , i.e set of limit points.
Conversely . Let $F$ be closed and $F\setminus\{x\}$ is closed. Then clearly $x$ is not a limit point of $F$ as if it was then it would also be a limit point of $x$ as
for every open nbd $U$ of $x$ , $U\cap F\setminus\{x\}= U\cap ((F\setminus\{x\})\setminus\{x\})\neq \phi$ . Hence $F\setminus\{x\}$ is a closed set which does not contain $x$ which is a limit point of it. Contradiction. Hence $x$ is not a limit point. So $x$ is an isolated point of the set.
