Say two subsets of $\mathbb{R}$ are equivalent if they are homeomorphic, with the subspace topology. How many equivalence classes are there?
It's immediate that there are at least $\beth_0$ (we can get any finite discrete topology) and no more than $\beth_2 = \mathcal{P}(\mathbb{R})$. I have a semi-complex construction below to show there are at least $\beth_1$.
Without the generalized continuum hypothesis, this question seems harder and possibly independent. I would be interested in an answer which assumes the continuum hypothesis, and thus answers either $\beth_1$ or $\beth_2$.
Let's not complicated things too much, but we could also ask this question on $S^1$ and $\mathbb{R}^2$: on $S^1$ the construction below does not work, and on $\mathbb{R}^2$ it's much easier.
Construction of $\beth_1$ non-homeomorphic subsets: The idea is that being a limit point, or being a limit point of limit points, or being a limit point of limit points of limit points... are topological properties. For a given set $A \subset \mathbb{R}$, call a point
- A $P_0$ point if it is in $A$.
- A $P_n$ point if it is the limit point of $P_{n-1}$ points.
- A $Q_n$ point if it is a $P_n$ point, but not a $P_{n+1}$ point.
- A $R_n$ point if it is a $Q_n$ point and each neighborhood of the point (in A) is uncountable.
The existence of a $R_n$ point is a topological invariant. I claim that for any $J \subset \mathbb{N}$, it's possible to create $A \subset \mathbb{R}$ which has a $R_n$ point if and only if $n \in J$.
To motivate the construction and the definition of $Q_n$, first think about constructing a $P_n$ point. For example, the set $\{0\} \cup \{ 1/k : k \in \mathbb{N} \}$ has $0$ as a $P_1$ point. Now, add some points to that set: on each interval $(1/(k+1), 1/k)$, add a sequence of points limiting to $1/(k+1)$. Then
- The points we just added are $P_0$ points.
- For each $k$, the point $1/k$ is a $P_1$ point and a $P_0$ point, and thus a $Q_1$ point but not a $Q_0$ point.
- 0 is a $P_2$ point, a $P_1$ point, and a $P_0$ point. Therefore it is a $Q_2$ point.
Now, it would be simpler if we didn't need the $R_n$ definition. But as you see, we can't make a $Q_n$ point without also making a $Q_{n-1}$ point. Now consider adding the interval $(-1, 0)$ to the set we've constructed so far. The $0$ would be a $R_2$ point, but there would be no $R_1$ or $R_0$ points.
To construct a set corresponding to $J \subset \mathbb{N}$, do similar constructions as above all around $\mathbb{R}$, for each $n \in J$.
Since every continuous function $f:X \rightarrow \mathbb{R}$ can be extended to a continuous function $f':G \rightarrow \mathbb{R}$ on a $G_{\delta}$ set, the equivalence class of every $X \subseteq \mathbb{R}$ has size at most continuum. So there must be $2^{2^{\omega}} = \beth_2$ many equivalence classes.