Every closed subset of $\mathbb{R}$ is a derived set How can I show that given a closed subset $A \subset \mathbb{R}$, there is some set $B \subset \mathbb{R}$ such that $A= B'$, where $B'$ denotes the derived set of $B$? I feel as though this should be simple, so perhaps I am missing something...
 A: Let's decompose $A$ into $A=L\cup I$ where $L=\{\text{$x\in A$|$x$ is a limit point of $A$}\}$ and $I=\{\text{$x\in A$|$x$ is an isolated point of $A$}\}$.
For each $x\in I$, there exist $r_x>0$ such that $B(x,r_x)$ contains no other points of $A$. Let 
$$S_x=\{x,x+\frac{r_x}2,x+\frac{r_x}3,... \}$$
Observe that by letting
$
S=\bigcup_{x\in I}S_x
$
, $S'=I$ (Prove this!). By construction $S$ is disjoint from $L$. Letting
$$ B=L\cup S$$
we can see that $B'=A$, indeed.
A: The idea is this: Since $A$ is closed, we know that $A'\subseteq A$. So, intuitively, we want $B$ to be $A$ together with "extra stuff," $E$. This extra stuff should have two properties:


*

*Every point of $A\setminus A'$ is a limit of points in $E$, but

*no point not in $A$ is a limit of points in $E\cup A$.
If we can find such an $E$, then $B=A\cup E$ will be as desired.

So our first question is:

What does $A\setminus A'$ look like?

As it turns out, there's a lot we can say about this set; but for now, we just need:


*

*The points in $A\setminus A'$ are (strongly) isolated: we may associate intervals $I_x$ to each $x\in A\setminus A'$ such that $I_x\cap A=\{x\}$ and $I_x\cap I_y=\emptyset$ for $x\not=y$.


EXERCISE 1: Prove that. (It may be helpful to first show that $A\setminus A'$ is countable . . . )
EXERCISE 2: For each $x\in A\setminus A'$, let $I_x$ be such an open interval. Since $I_x$ is open, we can find a sequence of points $x_0<x_1<x_2< . . . $, $x_i\in I_x$, with $\lim x_i=x$. Let $E=\{x_i: x\in A\setminus A'\}$. Show that $(E\cup A)'=A$.
