# Uncountably many non-homeomorphic compact subsets of the circle

As the title says, the question is whether there are uncountably many non-homeomorphic compact subsets of the unit circle.

I'm assuming this is true, but I wouldn't mind an elegant proof.

• Note that with the exception of the circle itself, you can think of any other subset as a subset of $[0,1]$. – Asaf Karagila Nov 18 '13 at 19:00
• True. But I still don't see how to immediately get the answer. I could play with the number of connected components, but that's probably not enough. So one needs to look at more complicated sets, and I have not enough imagination, I guess. – Martin Argerami Nov 18 '13 at 19:02

For each countable ordinal $\alpha$ let $X_\alpha=\omega^\alpha+1$, where the exponentiation is ordinal exponentiation, and $X_\alpha$ is the space of ordinals less than or equal to $\omega^\alpha$ with the order topology. This is a compact space, and $X_\alpha$ and $X_\beta$ have different Cantor-Bendixson ranks when $\alpha\ne\beta$, so $\{X_\alpha:\alpha<\omega_1\}$ is an uncountable family of pairwise non-homeomorphic countable compact Hausdorff spaces. It’s well-known that every countable ordinal space embeds in $\Bbb R$, hence in $\left[0,\frac12\right)$, and hence in the circle.
Added: Let $C$ be the middle-thirds Cantor set, and let $\{p_n:n\in\omega\}$ be an enumeration of the left endpoints of the deleted intervals. If $A\subseteq[\omega_1]^\omega$, let $A=\{\alpha_n:n\in\omega\}$ be the increasing enumeration, and attach a copy of $X_{\alpha_n}$ to $C$ at $p_n$ by identifying $\omega^{\alpha_n}$ with $p_n$; the rest of $X_{\alpha_n}$ should lie in the interval whose left endpoint is $p_n$. Call the resulting space $C_A$. The spaces $C_A$ for $A\in[\omega_1]^\omega$ are pairwise non-homeomorphic and compact, and there are $2^\omega$ of them.
• Nice, very nice. Can you get all the way up to $2^{\aleph_0}$ without $\sf CH$? – Asaf Karagila Nov 18 '13 at 19:07