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Are there uncountably many subsets of the Cantor set such that they are not homeomorphic to each other?

Motivation: Let $X$ be the space of infinite binary tree. Then $Ends(X):=\varprojlim_{K,\text{compact}}\pi_0(X-K)$ has to topology of Cantor set. Let $\Sigma_X$ be a surface which handles are glued to a $2$-sphere along the tree $X$. Then $Ends(X)=Ends(\Sigma_X)$ has the topology of Cantor set.

If there are uncountably many subsets of the Cantor set such that they are not homeomorphic to each other, then I can construct an uncountably infinite family of surfaces $\{\Sigma_{X'}\}_{X'\subseteq X}$ such that they are pairwise non-homeomorphic, and each is obtained by gluing countably many handles.

Y.H. Chan
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    Yes, using a similar argument in Are there more than $\beth_1$ non-homeomorphic topological subspaces of $\Bbb R$?. (The crux is Cantor set is separable; see the comments under Eric Wofsey's answer) – YuiTo Cheng May 14 '19 at 06:16
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    It is rather unclear to me how you intend to get your uncountable family of surfaces from this, though. How would you get a surface correspond to an arbitrary subset of $Ends(X)$? – Eric Wofsey May 14 '19 at 06:36
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    @YuiToCheng. That does answer the Q! I was going to post a long answer showing that (1) there is an uncountable set of countable ordinals that (with the $\in$-order topology) are not homeomorphic to each other, and (2) every countable ordinal is homeomorphic to a sub-space of the Cantor set. – DanielWainfleet May 14 '19 at 09:53
  • @DanielWainfleet Interesting approach... Aside: your (2) reminds me of the result that every countable compact Hausdorff space is homeomorphic to a successor ordinal. There are several proofs on this site and I think the simplest is https://math.stackexchange.com/questions/38247/countable-compact-spaces-as-ordinals/416672#416672, though I'm still wondering whether it's correct or not. – YuiTo Cheng May 14 '19 at 10:57
  • @EricWofsey Any (infinite) subgraph of the infinite binary graph $X$ corresponds to a subset of the Cantor set. This correspondence is 1-1 onto. To construct the surface, start with a tiny $2$-sphere at the "origin". Then glue handles along the edges of the subgraph. – Y.H. Chan May 14 '19 at 16:48
  • That correspondence is definitely not onto! There are only $\beth_1$ subgraphs of the infinite binary graph (since it's countable) but there are $\beth_2$ subsets of the Cantor set. – Eric Wofsey May 14 '19 at 16:49
  • You haven't said exactly how the correspondence is defined, but I would guess that it only hits the closed subsets of the Cantor set. – Eric Wofsey May 14 '19 at 16:50
  • @YuiToCheng . On the other hand the method of the linked Q gives a lot more spaces ($beth(2)$) and with little effort. I intend to look at that result about countable compact $T_2$ spaces. I do know how to show that such a space (if non-empty) is the continuous image of a successor ordinal. – DanielWainfleet May 14 '19 at 18:54
  • @EricWofsey The correspondence is constructed as follow: each element of the cantor set correspond to an infinite binary string. $0$ correspond to choosing the left one-third and $1$ correspond to choosing the right one-third. Similarly, each infinite ray of the infinite binary graph corresponds to an infinite binary string. So any subgroup of the Cantor set corresponds to some collection of rays, which is a subgraph. Vice versa. – Y.H. Chan May 15 '19 at 07:51
  • Different collections of rays can form the same subgraph! For instance, if you take all the rays except for one, that still forms the full binary tree, since every vertex of the tree is in some ray besides the one you omitted. – Eric Wofsey May 15 '19 at 15:06

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