# How many analytic sets does the Cantor set have?

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?

• 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... – Nate Eldredge Jul 18 '19 at 23:33
• Thank you! I didn't remember that characterization. It makes the problem much easier. – 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 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?
• 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. – Nate Eldredge Jul 18 '19 at 22:48