Willard 17R; uncountably many compact subsets of real line

Question

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).

Answer 1

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.

Answer 2

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 $[0,1]$.