How many closed subsets of $\mathbb R$ are there up to homeomorphism?

Solution 1:

Classifying all of the closed subsets of $\mathbb{R}$ up to homeomorphism seems quite hard, though perhaps not entirely intractible. However, it is not too hard to count how many of them there are: there are exactly $2^{\aleph_0}$ closed subsets of $\mathbb{R}$, up to homeomorphism.

First, since $\mathbb{R}$ is second-countable, there are only $2^{\aleph_0}$ closed subsets of $\mathbb{R}$, so there are at most $2^{\aleph_0}$ different closed subsets up to homeomorphism. So it suffices to give a family of $2^{\aleph_0}$ non-homeomorphic closed subsets of $\mathbb{R}$.

To construct such a family, let $f:\omega^\omega+1\to \mathbb{R}$ be a continuous order-preserving map from the ordinal $\omega^\omega+1$ to $\mathbb{R}$ (it's not too hard to construct such a map, and in fact it is not hard to prove by induction that for any countable ordinal $\alpha$, there is a continuous order-preserving map $\alpha\to\mathbb{R}$). For each $\alpha<\omega^\omega$, let $I_\alpha$ be the interval $[f(\alpha),(f(\alpha)+f(\alpha+1))/2]$. Say that an ordinal $\alpha<\omega^\omega$ has rank $n$ if it is of the form $\alpha=\omega^Nk_N+\omega^{N-1}k_{N-1}+\dots+\omega^nk_n$ for some $N,k_N,k_{N-1},\dots,k_n\in\mathbb{N}$ with $k_n\neq 0$. That is, the rank is the smallest power of $\omega$ appearing in the Cantor normal form of $\alpha$ (it is also the Cantor-Bendixson rank of $\alpha$ as an element of the space $\omega^\omega+1$).

Now if $A\subseteq\mathbb{N}$, define $$S_A=f(\omega^\omega+1)\cup\bigcup_{\operatorname{rank}(\alpha)\in A}I_\alpha.$$ Less formally speaking, we construct $S_A$ by taking the image of the map $f$ and adding a small interval to the right of every point whose rank is in $A$. It is easy to see that each $S_A$ is closed; I claim that we can recover the set $A$ from the homeomorphism type of $S_A$, so $S_A\cong S_B$ implies $A=B$. Indeed, consider the quotient space of $S_A$ obtained by collapsing each connected component of $S_A$ to a point. It is easy to see that this quotient space can be identified with $\omega^\omega+1$, with the quotient map $q:S_A\to\omega^\omega+1$ given by the inverse of $f$ on $f(\omega^\omega+1)$ and $q(x)=\alpha$ if $x\in I_\alpha$. Thus the preimage of an ordinal $\alpha$ under the map $q$ is a just $\{f(\alpha)\}$ if $\operatorname{rank}(\alpha)\not\in A$, and is the entire interval $I_\alpha$ if $\operatorname{rank}(\alpha)\in A$. Thus a natural number $n$ is in $A$ iff there exists a point of rank $n$ in $\omega^\omega+1$ whose inverse image under $q$ has more than one point. Since the rank of an element of $\omega^\omega+1$ can be defined purely topologically (as the Cantor-Bendixson rank), this is a description of the set $A$ using only the topological structure of $S_A$.

Thus for each subset $A$ of $\mathbb{N}$, we have given a closed subset $S_A$ of $\mathbb{R}$, such that different sets give non-homeomorphic closed subsets. Thus there are $2^{\aleph_0}$ non-homeomorphic closed subsets of $\mathbb{R}$.

Solution 2:

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 $A(X)=\{n\}.$