An example of set which is not Souslin set

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.

• Every Borel set is Suslin. And consistently, every set of reals is Borel. So you need to rely on the axiom of choice here. – Asaf Karagila Dec 11 '17 at 6:57
• @AsafKaragila I believe that "every Borel set is Suslin" depends on having codes for Borel sets, while "every set of reals is Borel" depends on not requiring Borel sets to have codes. So I don't think you can combine the two, because they use different definitions of "Borel". In fact, with the OP's definition of "Suslin set", it seems there are at most $2^{\aleph_0}$ Suslin sets, so some set of reals is not Suslin. (Probably, the set of codes of well-orderings will be an example in ZF, just as in the presence of AC.) – Andreas Blass Dec 16 '17 at 14:39
• In my previous comment, "at most $2^{\aleph_0}$ wasn't good wording. What I should have said is that the continuum can be mapped onto the collection of all Suslin sets, so that collection can't be the whole power set of the continuum. – Andreas Blass Dec 16 '17 at 15:39
• To clarify Asaf's comment for the OP: it is consistent with ZF that every set of reals is Borel. In ZFC - even with a small amount of choice - this is of course false. Moreover, there are simple examples (e.g. the set of reals coding well-orderings), so this isn't a case of choice letting you construct pathological sets but rather giving you the right tools to analyze the sets you already have. Reflecting this, there is a notion of "coded Borel set" which behaves much better in ZF, and indeed ZF alone proves that there are non-coded Borel sets. And Andreas' comment gets at the subtlety here. – Noah Schweber Dec 16 '17 at 16:06
• @Andreas: Good point. – Asaf Karagila Dec 16 '17 at 16:22

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.

• You probably should indicate why all Suslin sets are analytic. – Andrés E. Caicedo Dec 16 '17 at 16:00
• @AndrésE.Caicedo Oh, thanks. I'll fix it. – Pedro Sánchez Terraf Dec 16 '17 at 16:03
• Thank you very much for your answer – Maurten Erik Dec 17 '17 at 8:15
• @MaurtenErik Thank you for the reward! – Pedro Sánchez Terraf Dec 18 '17 at 12:55