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 $\mathbb{R}$?

  • $\begingroup$ I believe it may be possible to classify all of them, but the classification will be very complicated. In particular, there are uncountably many different closed sets up to homeomorphism. $\endgroup$ Jan 7, 2016 at 8:52
  • $\begingroup$ Has there been any results in this particular area? Where would be a good place to find them? $\endgroup$ Jan 7, 2016 at 8:54
  • $\begingroup$ Also, would it be good to ask this on overflow as well? $\endgroup$ Jan 7, 2016 at 9:03
  • $\begingroup$ @bof would you mind elaborating on that? $\endgroup$ Jan 7, 2016 at 9:14
  • $\begingroup$ @bof: That isn't quite right; some countable ordinals are homeomorphic to each other. However, the countable ordinals of the form $\omega^\alpha$ are pairwise non-homeomorphic. $\endgroup$ Jan 7, 2016 at 9:34

2 Answers 2


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 $\mathbb{R}$.

  • $\begingroup$ Can't you get $2^{\aleph_0}$ non-homeomorphic nowhere dense closed subsets of $\mathbb R$ by using Cantor sets instead of intervals? $\endgroup$
    – bof
    Jan 7, 2016 at 10:20
  • $\begingroup$ For any two Cantor sets $C_1,C_2 \subset \mathbb{R}$ there exists a homeomorphism $f : \mathbb{R} \to \mathbb{R}$ such that $f(C_1)=C_2$. So no, you cannot get that @bof. $\endgroup$
    – Lee Mosher
    Jan 7, 2016 at 15:55
  • $\begingroup$ @LeeMosher: bof was suggesting replacing the intervals $I_\alpha$ with Cantor sets in my construction. It is not obvious then whether the spaces $S_A$ would still be non-homeomorphic, however, because it seems that the quotient map $S_A\to \omega^\omega+1$ cannot be canonically defined. $\endgroup$ Jan 7, 2016 at 19:02
  • $\begingroup$ Is there anywhere that I can find sources to understand this answer? I don't want to accept something that I don't understand. $\endgroup$ Jan 8, 2016 at 2:41
  • 1
    $\begingroup$ What parts in particular don't you understand? (Also, you may find it easier to understand bof's answer instead, as it is somewhat simpler.) $\endgroup$ Jan 8, 2016 at 2:50

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


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