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 '16 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 '16 at 8:54
  • $\begingroup$ Also, would it be good to ask this on overflow as well? $\endgroup$ Jan 7 '16 at 9:03
  • $\begingroup$ @bof would you mind elaborating on that? $\endgroup$ Jan 7 '16 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 '16 at 9:34

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 '16 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 '16 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 '16 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 '16 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 '16 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 $A(X)=\{n\}.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.