TITLE: Are there more than $\beth_1$ non-homeomorphic topological subspaces of $\Bbb R$?
QUESTION [8 upvotes]: 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.

REPLY [10 votes]: 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 $\mathbb{R}$.