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I've been asked by a younger student about a certain claim he had on a classification of topological subsets of $\Bbb R$. The overall idea was a bit fuzzy, but in hindsight it revolved around taking the $\sigma$-algebra generated by six (Borel) subsets + translations. I successfully (and, I hope, instructively) argumented against it. However, this led me to the question:

Could I just cut it short and fancy with a cardinality argument? Specifically, if $\sim$ is the homeomorphism equivalence on $\mathcal P(\Bbb R)$, is $\operatorname{card}\left(\mathcal P(\Bbb R)/\sim\right)>\beth_1$ ?

Intuitively, I'd say yes, because, "come on, there are $\beth_2$ nasty non-Borel sets". And, "at chit-chat level, homeomorphisms $(a,b)\to(c,d)$ are monotone functions". However, this is neither a proof nor a sufficient reason for my question to even be decidable in ZFC.

In fact, on the topic I found this weaker fact: "closed subsets up to homeomorphism are exactly $\beth_1$".

Thank you for links and/or answers.

Andrés E. Caicedo
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1 Answers1

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Every subset of $\mathbb{R}$ has a countable dense subset. If $X\subseteq\mathbb{R}$ and $A\subseteq X$ is a countable dense subset, a homeomorphism from $X$ to another subset $Y\subseteq\mathbb{R}$ is determined by its restriction to $A$. So there is an injection from the set of homeomorphisms from $X$ to other subsets of $\mathbb{R}$ to the set of functions from $A$ to $\mathbb{R}$. There are only $\beth_1$ functions from $A$ to $\mathbb{R}$ since $A$ is countable.

So each subset of $\mathbb{R}$ can be homeomorphic to at most $\beth_1$ other subsets of $\mathbb{R}$. Since there are $\beth_2>\beth_1$ different subsets of $\mathbb{R}$, there must be $\beth_2$ different homeomorphism classes of subsets of $\mathbb{R}$.

Eric Wofsey
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  • Oh, in fact that wasn't hard at all! :) –  Feb 06 '17 at 19:26
  • So, I guess the same goes for any separable topological space with $\beth_1$ elements. –  Feb 06 '17 at 19:28
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    @G.Sassatelli: Not quite: you need every subspace to be separable as well (so, for instance, it suffices to assume your space is second-countable). You also need your space to be Hausdorff, so that a continuous function is determined by what it does on a dense subset. – Eric Wofsey Feb 06 '17 at 19:31
  • Thank you very much for the edit of my question – Sebastiano Aug 28 '20 at 22:22