Q:

Hausdorff space (derived set)

Prove that in Hausdorff spaces:

$A'$ is always closed.
$(A')' \subset A'$
The closure of $(A')=A'$

I have the following definition for $A'$ but I don't how to use it.
$$A'= \{ x \in X : \forall\text{ open } U \subseteq X\text{ such that }x\in U, ((U\setminus\{x\}) \cap A\ne \emptyset) \}$$
or is there another simpler definition? I know that the points of a Hausdorff space are closed, so are all the points of $A'$ closed too?
Thanks for your help.

A:

Let $X$ be a Hausdorff space and $A\subseteq X$ any nonempty subset. I will use your definition of $A^{\prime}$.
${\bf (i)}$ $A^{\prime}$ is closed.
To show that $A^{\prime}$ is closed it will suffice to show that $X\setminus A^{\prime}$ is open. Then let $y\in X\setminus A^{\prime}$. Then, by definition, there is an open neighbourhood $U\subseteq X$ that contains $y$ and is such that $(U\setminus\{y\})\cap A=\emptyset$. We then also have that $z\notin A^{\prime}$ for every $z\in U$ because $V:=U\setminus\{y\}$ is an open set containing $z$ and is such that $(V\setminus\{z\})\cap A=\emptyset$. Thus $U$ is completely contained in $X\setminus A^{\prime}$, which is to say that $X\setminus A^{\prime}$ is open and $A^{\prime}$ is closed by virtue of being the complement of an open set.
${\bf (ii)}$ $(A^{\prime})^{\prime}\subseteq A^{\prime}$
Let $x\in (A^{\prime})^{\prime}$ and let $U\subseteq X$ be an open set containing $x$. Then $(U\setminus\{x\})\cap A^{\prime}\neq\emptyset$. Then let $y\in (U\setminus\{x\})\cap A^{\prime}$. Because $V:=U\setminus\{x\}$ is open and contains $y$ we have that $(V\setminus\{y\})\cap A\neq\emptyset$. Thus we have that $(U\setminus\{x\})\cap A\neq\emptyset$ and $x\in A^{\prime}$. Therefore $(A^{\prime})^{\prime}\subseteq A^{\prime}$.
${\bf (iii)}$ The closure of $A^{\prime}$ is $A^{\prime}$.
I'm not sure what you are using as your definition for the closure of a set, but usually the closure of a set $B$ is defined to be the intersection of all closed sets containing $B$. In this case we proved in ${\bf (i)}$ that $A^{\prime}$ is closed, so the intersection of all closed sets containing $A^{\prime}$ is the intersection of a collection of sets that all contain $A^{\prime}$, including $A^{\prime}$ itself, so the intersection of this collection is $A^{\prime}$ trivially. In general, the closure of a closed set is always itself.
${\bf Exercise:}$ Where did I invoke the Hausdorff condition?

