## Counting submanifolds of the plane

After thinking about this question and reading this one I am led to ask for an uncountable collection of homeomorphism types of boundaryless connected path-connected submanifolds of the plane.

My guess is that is suffices to consider complements of Cantor sets. However, I do not know how to distinguish ends (up to homeomorphism) sufficiently well to ensure that this works. Are there other, easier, invariants?

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 Given two perfect, totally disconnected subsets $A,B$ of $\mathbb R^2$, is there always a homeomorphism $\varphi$ of $\mathbb R^2$ such that $\varphi(A)=B$? If the answer is yes, then it is not enough to work with complements of Cantor sets. – Roland Bacher May 17 2010 at 14:29

## 1 Answer

See a theorem of Richards, which implies that homeomorphism types of planar surfaces are in 1-1 correspondence with homeo. types of compact subsets of the Cantor set. I think there should be uncountably many homeo. types of totally disconnected compactums, but I don't know a reference or an argument off the top of my head. I think this should be related to the ordinalities of the accumulation points, but I'm not sure which ordinals can occur.

Addendum: Googling, I found references to a result of Markiewicz-Sierpinski classifying countable compact metric spaces up to homeomorphism by their Cantor-Bendixson rank (see section 3 of this paper for a statement). The CB-rank must be a countable ordinal $\zeta$, and the space is homeomorphic to the ordinal $\omega^\zeta\cdot n+1$ with the order topology for some $n\in \mathbb{N}$. These may all be realized as compact subsets of the line. This gives uncountably many non-homeomorphic compacta, which by Richards' theorem implies that there are uncountably many planar surfaces.

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 Very nice! I'll take a look at the paper you mention. – Sam Nead May 18 2010 at 10:59