# Every Suslin set is the continuous image of some closed set in $\mathbb N^{\omega}$.

Let $A$ be some countable set, and let $A^{\omega}$ be the set of all sequences $\mathbb N \to A$ equipped with the product topology. A subset $X \subseteq A^{\omega}$ is called Suslin set if there is a countable set $B$, a Borel subset $Y \subseteq B^{\omega}$ and a continuous map $f : B^{\omega} \to A^{\omega}$ such that $X = f(Y)$, i.e. it is the continuous image of some Borel set.

Does anybody has a proof of the following result:

If $X \subseteq A^{\omega}$ is a Suslin set, then there exists a closed set $Y \subseteq \mathbb N^{\omega}$ and a continuous map $f : \mathbb N^{\omega} \to A^{\omega}$ such that $X = f(Y)$.

It is mentioned in a book, but its proof is faulty and I cannot find it anywhere else.

As the countable sets are all equipped with discrete topology, might assume they are all $\omega$. So the problem is reduced to the following: if $Y\subset \omega^\omega$ is Borel, then there exists a continuous function $f$ (in fact continuous bijection) whose domain is a closed subset $F$ of $\omega^\omega$ and $f[F]=Y$ (because then you can compose two maps). You only need to verify the following $$\mathcal{A}=\{Y\subset \omega^\omega: \exists\text{ continuous bijection } f\ \exists\text{ closed }F\subset\omega^\omega \operatorname{dom}(f)=F, f''F=Y\}$$ contains a $\sigma$-algebra containing open sets (i.e. the sub algebra that is closed under taking complements). These sets are called Lusin.
• $\mathcal{A}$ is closed under countable intersection and countable disjoint union
• $\mathcal{C}=\{A\in \mathcal{A}: A^c\in \mathcal{A}\}$. Show $C$ is a $\sigma$-algebra containing open sets.
• I use “countable” for “bijective with a subset of $\omega$”. If the OP uses this notation the same way as me, you might not replace all countable sets by $\omega$. – Pedro Sánchez Terraf Aug 28 '16 at 23:32
• Why you just considered continuous functions $f : F \to \omega^{\omega}$? I asked for functions $f : \omega^{\omega} \to \omega^{\omega}$ with a closed $F \subseteq \omega^{\omega}$ such that $f(F) = Y$? Could every function $f : F \to \omega^{\omega}$ extended to a function on the whole $\omega^{\omega}$? – StefanH Aug 29 '16 at 12:09
• Yes. Each closed subset $A$ is a retract of $\omega^\omega$, that means there is a continuous function $f: \omega^\omega \to \omega^\omega$ such that $f''\omega^\omega=A$ and $f\restriction A = id_A$. However, I think the proof strategy should work if you delete $dom(f)=F$ in the definition of $\mathcal{A}$. – Jing Zhang Aug 29 '16 at 14:57
• Okay, I tried to prove that $\mathcal A$ is closed under countable intersection and union. So suppose $\{A_i\}_{i\in I} \in \mathcal A$ is some countable collection, then I have to show that $\bigcup_{i\in I} A_i \in \mathcal A, \bigcap_{i\in I} A_i \in \mathcal A$. I know that $A_i = f(F_i)$ for $f : F_i \to \omega^{\omega}$ continuous and $F_i$ closed. Now I have to find $F, F'$ closed and $f$ and $g$ such that $f(F) = \bigcap_{i\in I} A_i, g(F') = \bigcup_{i\in I} A_i$. Is it easy to construct those function, do you mind giving a hint? – StefanH Aug 31 '16 at 12:21