TITLE: How many closed subsets of $\mathbb R$ are there up to homeomorphism?
QUESTION [11 upvotes]: I know there are lists of convex subsets of $\mathbb{R}$ up to homeomorphism, and closed convex subsets of $\mathbb{R}^2$ up to homeomorphism, but what about just closed subsets in general of $\mathbb{R}$?

REPLY [3 votes]: Inasmuch as there are just $2^{\aleph_0}$ closed subsets of $\mathbb R$ all told, it will suffice to exhibit $2^{\aleph_0}$ nonhomeomorphic nowhere dense closed subsets of $\mathbb R.$
For $S\subseteq\mathbb R$ and $n\in\omega$ let $S^{(n)}$ denote the $n^{\text{th}}$ Cantor-Bendixson derivative of $S,$ i.e., $S^{(0)}=S,\ S^{(1)}=S',\ S^{(n+1)}=(S^{(n)})'.$
For $X\subseteq\mathbb R$ let $A(X)$ denote the set of all positive integers $n$ for which there exists a relatively open set $U\subseteq X$ such that $S^{(n-1)}\cap U\ne S^{(n)}\cap U=S^{(n+1)}\cap U\ne\emptyset.$
It will suffice to show that, for every set $A$ of positive integers, there is a nowhere dense closed set $X\subseteq R$ with $A(X)=A;$ in fact, it will suffice to show this for a one-point set $A=\{n\}$ where $n$ is a positive integer.
Given a positive integer $n,$ construct a closed set $X\subseteq\mathbb R$ of order type $\omega^n+\varphi$ where $\varphi$ is the order type of the Cantor set; then $A(X)=\{n\}.$