An example of set which is not Souslin set

Question

DEFINITION

Let $X$ be a topological space. A set $B \subseteq X$ is called Souslin set if there is a family of closed sets $\{F_s|s \in \mathbb{N}^{<\mathbb{N}} \}$ in $X$ such that $$B=\bigcup_{\sigma \in \mathbb{N^N}} \bigcap_{n=1}^{\infty} F_{\sigma\upharpoonright n}$$

Question

I know that a countable intersection of closed sets or countable union of closed sets is Souslin set. I want to find an example of set which is not Souslin set. First, I assume that $A=[0,1) \subseteq \mathbb{R}$ is not Souslin set, but $A$ is $F_\sigma.$ Then, my assumption is not true. How to construct a set which is not Souslin set? Any hint would be appreciated. Thank you very much.

Answer 1

You can find a proof of this in Kechris' Classical Descriptive Set Theory, Theorem 14.2 in combination with Theorem 25.7.

Let $\mathcal{N}$ be $\mathbb{N}^\mathbb{N}$ construed as the topological product of countably many copies of $\mathbb{N}$ with the discrete topology. Theorem 25.7 shows that a subset of $\mathcal{N}$ is Souslin if and only if it is analytic, i.e., the continuous image of a separable completely metrizable space.

Now Theorem 14.2 states that there exists an analytic set $A\subseteq\mathcal{N}$ that is not Borel. In the course of the proof, it is shown that $\mathcal{N}\setminus A$ is not analytic, so this is a very sharp example: a non analytic set with analytic complement.

In any case, the cardinality argument by Andreas Blass above is extremely simple: Each analytic subset of $\mathcal{N}$ is determined by a countable family of closed sets. Since there are exactly $2^{\aleph_0}$ of the latter, you have $(2^{\aleph_0})^{\aleph_0} =2^{\aleph_0}$ analytic subsets of $\mathcal{N}$, hence there are $2^{2^{\aleph_0}}$ non analytic subsets.

Both arguments go through for every separable completely metrizable uncountable space.