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: $F-\{x\}$ closed in $F$ iff $\{x\}$ is open in $F$ iff $x$ is isolated in $F$. QED as $F-\{x\}$ closed implies in particular that it's closed in $F$ too.
A: One side
Assume $ F \setminus \{x\} $ is closed. This means that it's complement $ F^c \cup \{x\} $ is open. So there exists some $r_0 >0$ for which the neighborhood $ \mathcal{N}_{r_0}(x)$ is a subset of  $F^c \cup \{x\}$, meaning that this neighborhood of $x$ contains no element of $F$ except possibly $x$ itself (actually it contains $x$). So $x$ is an isolated point of $F$.

let $\overline{E}$ denote closing of $E$.
Other side
Assume that $x$ is an isolated point for $F$. To prove the closeness of $ F \setminus \{x\} $, you mentioned $F \setminus \{x\} \subset \overline{F \setminus \{x\}}$ is clear. For the other side of inclusion, first note that since $ F \setminus \{x\} \subset F $ we have $ \overline{F \setminus \{x\}} \subset \overline{F} $ (it's straight-forward, you can prove this if you haven't seen it before).
Now suppose $ y \in  \overline{F \setminus \{x\}}$. First we note that $y$ cannot be equal to $x$: Our $x$ was an isolated point for $F$, this by definition means that for some $ r > 0 $ the neighborhood $ \mathcal{N}_r(x) $, not considering $x$ itself, contains no member of $F$. But for $x$ to be in $ \overline{F \setminus \{x\}} $, there must have been some member of $ F \setminus \{x\} $ in any neighborhood of $x$, including $ \mathcal{N}_r(x) $. So $ y \neq x$.
Now suppose $ y \in \overline{F \setminus \{x\}} \subset  \overline{F} $ and by closeness of $F$ we have $ y \in F$, and since $ y \neq x $ we have $ y \in F \setminus \{x\} $.
A: 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.
