A set $S$ is called a derived set if there exists a set $T$ whose limit points are $S$.
Is there a classification of which subsets of $\mathbb{R}$ are derived sets?
It might be helpful to have a list of derived sets. Here are some examples:
- Any finite set.
- Any perfect set e.g. Cantor set
The derived sets are the same as the closed sets. Indeed, let $C\subseteq\mathbb{R}$ be closed. Then every non-isolated point of $C$ is a limit point of $C$. We just need to enlarge $C$ to make each isolated point a limit point. There are only countably many isolated points (each is isolated by some interval with rational endpoints), so let us enumerate them as $\{c_i:i\in\mathbb{N}\}$ (if there are only finitely many, the argument is only easier). Let $$D=C\cup\{c_i+1/2^j:i,j\in\mathbb{N}\text{ and }j>i\}.$$
Clearly every point of $C$ is a limit point of $D$. Conversely, I claim any limit point of $D$ is in $C$. Suppose $(x_n)$ is a sequence of distinct elements of $D$ with limit $x$. If infinitely many of the $x_n$ are in $C$, they form a subsequence whose limit $x$ must be in $C$ since $C$ is closed. So we may assume that $x_n\not\in C$ for all $n$. If infinitely many of the $x_n$ are of the form $c_i+1/2^j$ for some fixed $i$, then they form a subsequence which converges to $c_i$ so $x=c_i\in C$. So, we may assume that $x_n=c_{i_n}+1/2^{j_n}$ where $i_n\to\infty$. The requirement that $j>i$ in the definition of $D$ implies that $j_n\to\infty$ as well. But then $1/2^{j_n}\to 0$, so that the sequence $(c_{i_n})$ of elements of $C$ also converges to $x$, so $x\in C$ since $C$ is closed.
Conversely, it is easy to see any derived set is closed (if $x_n\to x$ and each $x_n$ is a limit point of $S$, then you can approximate each $x_n$ with elements of $S$ to get a sequence of elements of $S$ converging to $x$).