Definition. A Polish space is a separable completely metrizable topological space.

Definition. Let $X$ be a Polish space. A set $A\subseteq X$ is analytic if it is the continuous image of a Polish space.

I'm trying to show that the Cantor set has $\mathfrak{c}$ analytic sets. I could already show that $2^{\omega}$ has at least $\mathfrak{c}$ analytic sets.

Why does $2^{\omega}$ have at most $\mathfrak{c}$ analytic sets?

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    $\begingroup$ A convenient alternative characterization is that $A$ is analytic if and only if it is the projection of a closed subset of $X \times \omega^\omega$, where $\omega^\omega$ is Baire space. So this reduces the problem to counting the closed subsets of $2^\omega \times \omega^\omega$ which is itself a Polish space. And this in turn is equivalent to counting the open sets, which is not too hard because the space is second countable... $\endgroup$ – Nate Eldredge Jul 18 '19 at 23:33
  • $\begingroup$ Thank you! I didn't remember that characterization. It makes the problem much easier. $\endgroup$ – Fernando Mauricio Rivera Vega Jul 19 '19 at 17:43

Every analytic subset $A$ of a Polish space $X$ is determined by two things: another Polish space $Y$ and a continuous map $f:Y\rightarrow X$ (with $A=im(f)$).

So the natural thing to do, if we want to count analytic subets, is to count these descriptions: how many Polish spaces are there, and for a given pair of Polish spaces how many continuous maps between them are there? Of course an analytic set has multiple descriptions of the above type, but we can still get an upper bound - which, if it coincides with our existing lower bound, will answer the question.

Unsurprisingly the punchline is going to be that (up to isomorphism) there are continuum many Polish spaces, and for any pair of Polish spaces there are (at most) continuum many maps between them - so any Polish space has at most continuum many analytic sets.

So how do we do this counting? Well, let's think about counting continuous maps first. Suppose I have topological spaces $P$ and $Q$; what is a continuous map $h:P\rightarrow Q$ determined by? Well, it's determined by how it behaves on open sets: namely, it's determined by knowing which open sets in the domain it maps to which open sets in the range. And this can be simplified substantially: we can restrict attention to basic open sets, once we fix bases. Now Polish spaces are second-countable (being separable and metrizable), so can you see how to count the descriptions of continuous maps between two of them?

As to counting Polish spaces themselves, it's actually a bit easier to count complete separable metric spaces (again giving an upper bound). The trick is that if I know a dense subspace of a complete metric space, then I know the whole complete metric space - do you see how to formalize this, prove this, and use this to count complete separable metric spaces?

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    $\begingroup$ A somewhat related way to think about this is that every Polish space is homeomorphic to a closed subset of $\mathbb{R}^\omega$, which is itself a Polish space, and every continuous map between two such sets has a graph which is a closed subset of $(\mathbb{R}^\omega)^2$. So if you know that a Polish space has continuum many closed subsets, you are done. $\endgroup$ – Nate Eldredge Jul 18 '19 at 22:48

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