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).

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    $\begingroup$ 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 '20 at 20:05
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    $\begingroup$ 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 '20 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 '20 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 '20 at 20:14

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.

  • $\begingroup$ 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 '20 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 '20 at 1:41
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    $\begingroup$ You could just assume that the sequence $X_n$ contains each element of the countable collection twice. $\endgroup$ – bof Oct 20 '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 '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 '20 at 2:26

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 $[0,1]$.


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