5
$\begingroup$

I am self-studying topology and came across question 17R of Willard's General Topology.

17R. Compact subsets of $\mathbb{R}$
There are uncountably many nonhomeomorphic compact subsets of $\mathbb{R}$. [Use ordinals.]

The discussions I found which are similar (e.g. Uncountably many non-homeomorphic compact subsets of the circle) use what seems to be more advanced stuff ("Cantor-Bendixson rank", for example).

I guess the hint suggest us to look at $\Omega=[0,\omega_1]$, where $\omega_1$ is the first uncountable ordinal. I can do the following:

  • Every countable ordinal embeds into $\mathbb{R}$. This is more-or-less straightforward induction.

So the problem boils down in proving that there are uncountably many non-homeomorphic countable ordinals. It is also clear that if $\alpha$ is an infinite ordinal and $\beta$ is the largest limit ordinal $\leq\alpha$, then the compacts $[0,\alpha]$ and $[0,\beta]$ are homeomorphic.

I can also prove that there are uncountably many countable limit ordinals, but some of these are homeomorphic to each other (e.g. $\omega^2+\omega$ and $\omega^2$).

I would appreciate help, using not much more than basic facts about $\omega_1$ (as it is introduced in Willard's book).

$\endgroup$
  • 2
    $\begingroup$ It's not true that every infinite ordinal is homeomorphic to a limit ordinal. In fact, successor ordinals are never homeomorphic to limit ordinals, since successor ordinals are compact and nonzero limit ordinals never are. $\endgroup$ – Eric Wofsey Oct 19 at 20:05
  • 2
    $\begingroup$ In any case, I doubt there is a more elementary way to prove this than Cantor-Bendixson rank. But Cantor-Bendixson rank is pretty simple: if you're already familiar with basic pointset topology and transfinite induction, you should be able to understand it pretty easily. So I would suggest you just look it up and use it. $\endgroup$ – Eric Wofsey Oct 19 at 20:09
  • $\begingroup$ @EricWofsey Woops, you're right. What I was thinking was that "for every infinite ordinal $\alpha$, if $\beta$ is the largest limit ordinal $\leq\alpha$, then $[0,\alpha]$ and $[0,\beta]$ are homeomorphic. You are right and I will fix the question. $\endgroup$ – Questioner Oct 19 at 20:10
  • $\begingroup$ @EricWofsey On the second comment. I am starting to think that some of Willard's exercises are meant for us to go around and get to see more stuff on our own. I will definitely have a better look at Cantor-Bendixson. Thank you $\endgroup$ – Questioner Oct 19 at 20:14
8
$\begingroup$

Here is a proof that does not use ordinals at all. The crucial ingredient is the following lemma.

Lemma: Let $X$ be a nonempty compact Hausdorff space and suppose there exist two embeddings $f_0,f_1:X\to X$ with disjoint images. Then $X$ is uncountable.

Proof: The idea is that by iterating $f_0$ and $f_1$, you get a fractal of copies of $X$ similar to the Cantor set, which then must accumulate at uncountably different points. To make this precise, for any finite sequence $s$ of $0$s and $1$s, let $f_s$ be the corresponding composition of $f_0$s and $f_1$s. For any infinite sequence $r$ of $0$s and $1$s, let $X_r=\bigcap_s f_s(X)$ where $s$ ranges over all finite initial segments of $r$. Note that each $f_s(X)$ is a nonempty closed set, and they are nested, so by compactness, each $X_r$ is nonempty. But if $r\neq r'$, then $X_r$ and $X_{r'}$ are disjoint, since if you let $s$ and $s'$ be the first corresponding initial segments that differ, then $f_s(X)$ and $f_{s'}(X)$ are disjoint since $f_0$ and $f_1$ have disjoint images. Since there are uncountably many choices of $r$, this means $X$ be uncountable.

Theorem: There are uncountably many homeomorphism classes of countable compact subsets of $\mathbb{R}$.

Proof: Let $\{X_n:n\in\mathbb{N}\}$ be any countable collection of countable compact subsets of $\mathbb{R}$. Embed a copy of $X_n$ in the interval $(\frac{1}{n+1},\frac{1}{n})$ for each $n$, and let $Y\subset\mathbb{R}$ be the union of all these copies together with $0$. Finally, let $Z\subset\mathbb{R}$ be the union of two disjoint translated copies of $Y$. Then $Z$ is a countable compact subset of $\mathbb{R}$. However, each $X_n$ has two disjoint copies that embed in $Z$ (one in each copy of $Y$), so by the Lemma, $Z$ cannot be homeomorphic to $X_n$ for any $n$. Thus $\{X_n:n\in\mathbb{N}\}$ is not a complete list of all the countable compact subsets of $\mathbb{R}$ up to homeomorphism.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So you did better than the exercise wanted: You showed that there exist uncountably many nonhomeomorphic countable compact subsets of $\mathbb{R}$. Noice. $\endgroup$ – Questioner Oct 19 at 21:11
  • $\begingroup$ Nice. What about $\mathfrak c$ nonhomeomorphic compact subsets of $\mathbb R$? Can there be only $\aleph_1$ types of compact subsets while $\mathfrak c\gt\aleph_1$? $\endgroup$ – bof Oct 20 at 1:41
  • 1
    $\begingroup$ You could just assume that the sequence $X_n$ contains each element of the countable collection twice. $\endgroup$ – bof Oct 20 at 1:45
  • $\begingroup$ @bof: Hah, that was actually my original idea, but this way felt slightly easier to write up. $\endgroup$ – Eric Wofsey Oct 20 at 2:18
  • $\begingroup$ @bof: As for your question, both of us have in fact answered it before (the answer is there are always $\mathfrak{c}$ of them): math.stackexchange.com/questions/1602978/…. (OK, that was about closed subsets, but the constructions in my answer in fact gives compact subsets, and your answer can easily be arranged to do so.) $\endgroup$ – Eric Wofsey Oct 20 at 2:26
3
$\begingroup$

Using Eric Wofsey’s lemma we can explicitly construct an uncountable family of pairwise non-homeomorphic countable, compact spaces that embed in $\Bbb R$. They are the same ones that we’d get using a Cantor-Bendixson approach, but without all of that machinery.

If $X$ is a compact space that can be embedded in $[0,1]$, let $X^*$ be the one-point compactification of $\omega\times X$, where $\omega$ has the discrete topology; it’s not hard to show that $X^*$ can also be embedded in $[0,1]$. Use Eric Wofsey’s lemma to show that if $X$ is countable, then $X^*$ is not homeomorphic to $X$.

Now let $X_0$ be the compact ordinal space $\omega+1$. Given $X_\alpha$ for some ordinal $\alpha$, let $X_{\alpha+1}=(X_\alpha)^*$. If $\alpha$ is a countable limit ordinal, and $X_\eta$ has been defined for each $\eta<\alpha$, let $Y_\alpha=\bigsqcup_{\eta<\alpha}X_\eta$, and let $X_\alpha=(Y_\alpha)^*$; since $\alpha$ is countable, $X_\alpha$ can be embedded in $[0,1]$, and we can continue the recursive construction to get for each $\alpha<\omega_1$ a countable, compact space $X_\alpha$ that embeds in $[0,1]$.

Suppose that $\alpha<\beta<\omega_1$; then $X_{\alpha+1}\subseteq X_\beta$, and $X_{\alpha+1}$ contains disjoint copies of $X_\alpha$, so the lemma ensures that $X_\beta$ cannot be homeomorphic to $X_\alpha$, and $\{X_\alpha:\alpha<\omega_1\}$ is therefore an uncountable family of mutually non-homeomorphic countable, compact spaces that can be embedded in $[0,1]$.

| cite | improve this answer | |
$\endgroup$

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.