Derived subsets and topological spaces My question is regarding question $56$ from this GRE practice exam [PDF]:

For a subset $S$ of a topological space $X$, let $\operatorname{cl}(S)$ denote the closure of $S$ in $X$, and let $S'=\{x:x\in\operatorname{cl}(S\setminus\{x\})\}$ denote the derived set of $S$. If $A$ and $B$ are subsets of $X$, which of the following statements are true?
  
  
*
  
*$(A\cup B)'=A'\cup B'$  
  
*$(A\cap B)'=A'\cap B'$  
  
*If $A'$ is empty, then $A$ is closed in $X$.  
  
*If $A$ is open in $X$, then $A'$ is not empty.
  
  
  (A) $(1)$ and $(2)$ only
  (B) $(1)$ and $(3)$ only
  (C) $(2)$ and $(4)$ only
  (D) $(1)$, $(2)$, and $(3)$ only
  (E) $(1)$, $(2)$, $(3)$, and $(4)$

Intuitively I visualize the derived set $S'$ as the boundary of a set, although this seems insufficient. I am wondering if there is a good reference for a quick exposition of derived subsets. It would be great if the response can include a geometric description or understanding of $S'$ since this is how I usually like to see things in my head.
 A: Corrected to use the right definition of derived set.
The points of $S$ that are not in $S'$ are precisely the isolated points of $S$, i.e., the points $x\in S$ such that $\{x\}$ is a relatively open subset of $S$. In addition, $S'$ contains every limit point of $S$ that is not already in $S$. Thus, $S'$ consists of those points of $\operatorname{cl}S$ that are not isolated points in $\operatorname{cl}S$.
Thus, if $S'=\varnothing$, every point of $\operatorname{cl}S$ is isolated in $\operatorname{cl}S$, and $\operatorname{cl}S$ therefore has the discrete topology as a subset of $X$. But then $\operatorname{cl}S=S$ (why?), so $S$ is closed. This settles $(3)$: $(3)$ is true.
To settle $(4)$ we can use the often useful approach of looking at extreme cases: what if $A=\varnothing$? Then $A$ is certainly open, but $A'$ is by definition a subset of $A$, so $A'=\varnothing$, and we see that $(4)$ need not be true. If you want a more substantial example, take $X$ to be $\Bbb Z$ with the discrete topology, and let $A$ be any subset of $X$: $A$ is certainly open in $X$, but $A'=\varnothing$.
At this point you know that the only possible answers are (B) and (D), so $(1)$ is presumably true. If you have time later, you can come back and prove it.
Proof of $\mathbf{(1)}$:

Suppose that $x\in A'$. Then $$\begin{align*}x&\in\operatorname{cl}(A\setminus\{x\})\\&\subseteq\operatorname{cl}(A\setminus\{x\})\cup\operatorname{cl}(B\setminus\{x\})\\&=\operatorname{cl}\big((A\setminus\{x\})\cup(B\setminus\{x\})\big)\\&=\operatorname{cl}\big((A\cup B)\setminus\{x\}\big)\;,\end{align*}$$ so by definition $x\in(A\cup B)'$. Thus, $A'\subseteq(A\cup B)'$, and by symmetry $B'\subseteq(A\cup B)'$, so $A'\cup B'\subseteq(A\cup B)'$.
Now suppose that $x\in(A\cup B)'$. Then, making use of the previous calculations to skip a few steps, we have $$\begin{align*}x&\in\operatorname{cl}\big((A\cup B)\setminus\{x\}\big)\\&=\operatorname{cl}(A\setminus\{x\})\cup\operatorname{cl}(B\setminus\{x\})\end{align*}$$ and hence $x\in\operatorname{cl}(A\setminus\{x\})$ or $x\in\operatorname{cl}(B\setminus\{x\})$. But then $x\in A'$ or $x\in B'$, so $(A\cup B)'\subseteq A'\cup B'$. Combining results, we see that $(1)$ holds.

To choose between (B) and (D), we must see whether $(2)$ is necessarily true. And in fact it’s not: take $A=[0,1)$ and $B=(1,2]$ with $X=\Bbb R$. Then $(A\cap B)'=\varnothing$, but $A'\cap B'=[0,1]\cap[1,2]=\{1\}$. Thus, the correct answer is (B).
A: (4) is not correct, the discrete topology is a counter-example. (2) seems also not right, because for $X=\mathbb{R}$, $A=(-\infty,0)$ and $B=(0,\infty)$, $(A\cap B)'=\emptyset\neq A'\cap B'.$ This excludes all answers except (B).
