# Willard 17R; uncountably many compact subsets of real line

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).

• 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. – Eric Wofsey Oct 19 '20 at 20:05
• 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. – Eric Wofsey Oct 19 '20 at 20:09
• @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. – Questioner Oct 19 '20 at 20:10
• @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 – 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.

• So you did better than the exercise wanted: You showed that there exist uncountably many nonhomeomorphic countable compact subsets of $\mathbb{R}$. Noice. – Questioner Oct 19 '20 at 21:11
• Nice. What about $\mathfrak c$ nonhomeomorphic compact subsets of $\mathbb R$? Can there be only $\aleph_1$ types of compact subsets while $\mathfrak c\gt\aleph_1$? – bof Oct 20 '20 at 1:41
• You could just assume that the sequence $X_n$ contains each element of the countable collection twice. – bof Oct 20 '20 at 1:45
• @bof: Hah, that was actually my original idea, but this way felt slightly easier to write up. – Eric Wofsey Oct 20 '20 at 2:18
• @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.) – 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 $$[0,1]$$.