My question is more or less related to basic set theory. But I don't know even that. Apologies if I added the wrong tags.



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> **Motivation:** How many non-compact (planar) surfaces are there upto homeomorphism?
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> **Question:** How many pairwise non-homeomorphic non-empty **closed** subsets of the Cantor set are there?
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My idea is to produce an uncountable family $\mathcal F$ of **closed** subsets of the Cantor sets such that any two distinct elements of $\mathcal F$ are non-homeomorphic. Once I show this, the rest follows from the fact below:


*Let $\mathcal P\_1,\mathcal P\_2$ be two non-empty, **closed** subsets of the Cantor set. Then $\Bbb S^2\setminus \mathcal P\_1$ is homeomorphic to $\Bbb S^2\setminus \mathcal P\_2$ if and only if $\mathcal P\_1$ is homeomorphic to $\mathcal P\_2$. This is a very particular case of [Kerékjártó's classification theorem of non-compact surfaces](https://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf).*




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More generally, I have the following question:


Let $\mathcal F'$ be the collection of all pairs $(\mathcal P,\mathcal A)$, where $\mathcal P$ is a *non-empty closed* subset of the Cantor set, and $\mathcal A$ is a *closed* subset of $\mathcal P$. Define an equivalence relation $\sim$ on $\mathcal F'$ as follows: $(\mathcal P\_1, \mathcal A\_1)\sim (\mathcal P\_2, \mathcal A\_2)$ if and only if there is a homeomorphism $\varphi\colon \mathcal P\_1\to \mathcal P\_2$ with $\varphi(\mathcal A\_1)=\mathcal A\_2$.



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> **Question:** What's the cardinality of $\mathcal F'/\sim$?
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How many pairwise non-homeomorphic non-empty closed subsets of the Cantor set are there?
There are $2^{\aleph\_0}$ different subsets of the Cantor set up to homeomorphism.


There can't be more than $2^{\aleph\_0}$ of them because any subset of the Cantor set is separable. To construct $2^{\aleph\_0}$ of them, consider first the ordinal spaces $A\_n=\omega^n+1$, with their order topologies. All of them [can be imbedded in the Cantor set](https://math.stackexchange.com/questions/395931/is-any-compact-metric-totally-disconnected-space-homeomorphic-to-a-compact-subsp), and $A\_n$ has [Cantor Bendixon rank](https://proofwiki.org/wiki/Definition:Cantor-Bendixson_Rank) $n+1$.


Now, given any subset $S$ of $\mathbb{N}$, consider the usual, ternary Cantor set $C\subseteq[0,1]$ and the sequence $a\_n=1-\frac{1}{3^n}$, so that the intervals $I\_n=(a\_n-\frac{1}{3^n},a\_n]$ intersect $C$ only in $a\_n$. For each $n\in S$ we can imbed a copy of $A\_n$ in $I\_n$, with $\omega^n$ identified to the point $a\_n\in C$.


The resulting space $X\_S\subseteq[0,1]$ is compact (because all ordinals are compact and the intervals $I\_n$ converge to $1$). So as it is totally disconnected, it is homeomorphic to a closed subset of the Cantor set.


We just need to prove that $S$ can be obtained from the topology of $X\_S$, so that different subsets of $\mathbb{N}$ give rise to different spaces up to homeomorphism. To that end, consider the set $Y\_S\subseteq X\_S$ formed by points which are not condensation points, and let $Z\_S=\overline{Y\_S}$ be its closure. $Z\_S$ contains $Y\_S$, the points $a\_n$, with $n\in\mathbb{S}$, and no more points of $X\_S$ (except for $1$ if $S$ is infinite). Moreover, the Cantor Bendixon rank of $a\_n$ in $Z\_S$ is $n+1$, and if $1$ is in $Z\_S$ then its Cantor Bendixon rank is $\infty$, so yes, $S$ can be obtained from the Bendixon ranks of the points of $Z\_S\setminus Y\_S$.


