definition of isolated/limit point with closure

Question

The following is the definition of isolated point in my textbook:

Let $X$ be a topological space $X$, and let $A\subseteq X$. A point $x\in A$ is said to be an isolated point of $A$ if the one-point set $\{x\}$ is open in $A$.

Also, the following are some properties for limit and isolated points:

Theorem $x\in A$ is an isolated point if and only if $x\in A\setminus A'$, where $A'$ is the set of limit points of $A$.

Theorem Any closed subset of $X$ can be written as a disjoint union of its limit points and isolated points.

Intuitively, it seems to be "$x\in A$ is an isolated point if and only if $x\in\overline{A}\setminus A'$", where $\overline{A}$ is the closure of $A$.

Is there a counterexample so that $x\in\overline{A}\setminus A'$ is not an isolated point, or $x$ is an isolated point but $x\not\in\overline{A}\setminus A'$?

Give some examples! Thank you!

Answer 1

Points of $A$ are of two types, purely by logic:

• all open neighbourhoods of $x$ contain a point of $A$ not equal to $x$.

• there is an open neighbourhood of $x$ such that the only point of $A$ it contains is $x$.

The second type are the isolated points, by definition. The former are the limit points of $A$ (that are in $A$) so $A \cap A'$.

If $x \notin A$ it can only be of the first type, and then it's a point of $A'\setminus A$, and part of the closure of $A$.

So a point of $\overline{A} \setminus A'$ must be an isolated point of $A$ and all isolated points of $A$ are in $\overline{A} \setminus A'$.

Answer 2

There are no such counterexamples.

Hint:

The statement $$ \overline A= A\cup A' $$ is very helpful for your question.

Proof of the statement:

Let $x\in X\setminus \overline A$. Then there is an open set $O_x$ containing $x$ such that $O_x\subset X\setminus\overline A$. Thus $x$ is not in $A'$ and also not in $A$.

Conversely, let $x\in X\setminus (A\cup A')$. Then there is an open set $O_x$ containing $x$ such that $(O_x\setminus\{x\})\cap A=\emptyset$. Since $x\not\in A$, it follows that $O_x\cap A=\emptyset$. Since $O_x$ is open, it follows that $\overline A\subset X\setminus O_x$. Thus $x$ is not in $\overline A$.