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).
Best Answer
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.