We say that a subset of a topological space is Suslin if and only if it is the result of the Suslin operation applied on a Suslin scheme consisting of closed sets.
I know that for a Polish space $X$ the family of all Suslin subsets of $X$ is closed under countable unions and countable intersections. That follows from the fact that a subset of a Polish space is Suslin if and only if it is analytic (and it is well-known and quite easy to prove that countable union or intersection of analytic sets is analytic).
However, I don't know how to prove the closedness under countable unions or intersections in the more general case of all metrizable spaces (or even all topological spaces).
Does it even hold true?
I'll appreciate any help.
Best Answer
In Classical Descriptive Set Theory (A. Kechris) Thm. 25.6 states that if $X$ is a set and $\Gamma \subseteq \operatorname{Pow}(X)$, then $\mathcal{A}\mathcal{A}\Gamma = \mathcal{A}\Gamma$, where $\mathcal{A}\Gamma$ for a family of sets $\Gamma$ is the result of all possible applications of the Suslin (Kechris uses the French transliteration Souslin) operation on systems $(P_s)_{s \in \omega^{< \omega}}$ with all $P_s$ in $\Gamma$.
As countable unions and countable intersections of sets are special cases of the Suslin operation, the result follows, e.g. for all closed sets in topological spaces, as is your interest.
Fun fact: the notation $\mathcal{A}$ is in honour of Aleksandrov (Alexandroff), the Russian topology pioneer. The coincidence with "analytic" is a nice bonus.