Exercise 6, § 6

Let X be a metrizable space. (a) Now prove that the set of Borel subsets of X is the smallest subset $\mathscr{F}$ of $\mathfrak{P}(X)$ containing the closed sets of X and such that every countable union and every countable intersection of sets of $\mathscr{F}$ belong to $\mathscr{F}$. (Consider the subset $\mathfrak{G}$ of formed by the $A \in \mathscr{F}$ such that $X - A \in \mathscr{F}$). (b) Now prove that the set of Borel subsets of X is the smallest subset $\mathscr{F}$ of $\mathfrak{P}(X)$ containing the open sets of X and such that every countable intersection of sets of $\mathscr{F}$ belongs to $\mathscr{F}$, as well as every union of a sequence of sets of $\mathscr{F}$, pairwise disjoint (same method). (c) Every ordinal $\alpha$ can be written in a unique way $\alpha = \omega \beta + n$, where $n < \omega$ (E, III, p. 77, exerc. 14); one says that $\alpha$ is pair (resp. odd) if $n$ is pair (resp. odd). For every countable ordinal $\alpha$, one defines by transfinite induction the sets of subsets $\mathfrak{F}_\alpha(X)$, $\mathfrak{G}_\alpha(X)$ (or simply $\mathfrak{F}_\alpha$, $\mathfrak{G}_\alpha$) as follows: $\mathfrak{F}_0$ (resp. $\mathfrak{G}_0$) is the set of closed (resp. open) subsets of X; if $\alpha$ is pair and $> 0$, $\mathfrak{F}_\alpha$ (resp. $\mathfrak{G}_\alpha$) is the set of subsets of X which are countable intersections (resp. countable unions) of subsets belonging to $\bigcup_{\xi < \alpha} \mathfrak{F}_\xi$ (resp. to $\bigcup_{\xi < \alpha} \mathfrak{G}_\xi$); if $\alpha$ is odd, $\mathfrak{F}_\alpha$ (resp. $\mathfrak{G}_\alpha$) is the set of subsets of X which are countable unions (resp. countable intersections) of subsets belonging to $\bigcup_{\xi < \alpha} \mathfrak{F}_\xi$ (resp. to $\bigcup_{\xi < \alpha} \mathfrak{G}_\xi$). Now prove that the union of the $\mathfrak{F}_\alpha$ (resp. of the $\mathfrak{G}_\alpha$) is the set of Borel subsets of X (use a)). Deduce that if X is of countable type, the set of Borel subsets of X has a power at most equal to the power of the continuum. (d) The relation $A \in \mathfrak{F}_\alpha$ is equivalent to $\mathbf{C} A \in \mathfrak{G}_\alpha$. Every finite union (resp. intersection) of sets of $\mathfrak{F}_\alpha$ (resp. $\mathfrak{G}_\alpha$) belongs to $\mathfrak{F}_\alpha$ (resp. $\mathfrak{G}_\alpha$). We have $\mathfrak{F}_\alpha \subset \mathfrak{F}_{\alpha+1} \cap \mathfrak{G}_{\alpha+1}$ and $\mathfrak{G}_\alpha \subset \mathfrak{F}_{\alpha+1} \cap \mathfrak{G}_{\alpha+1}$ (reason by transfinite induction). (e) Let $f$ be a continuous mapping of X into a metrizable space Y. Now prove that

(f) If X and Y are two metrizable spaces, every product $A \times B$, where $A \in \mathfrak{F}_\alpha(X)$, $B \in \mathfrak{F}_\alpha(Y)$ (resp. $A \in \mathfrak{G}_\alpha(X)$, $B \in \mathfrak{G}_\alpha(Y)$) belongs to $\mathfrak{F}_\alpha(X \times Y)$ (resp. $\mathfrak{G}_\alpha(X \times Y)$). (g) Let $(X_n)$ be a sequence of metrizable spaces. Now prove that if, for every $n$, $A_n$ is Borel in $X_n$, then $\prod_n A_n$ is Borel in $\prod_n X_n$.

¹ It is not known whether every scattered metrizable space is strongly scattered.

h) Let $\alpha$ be a countable even ordinal (resp. odd), and let $(A_n)$ be a countable covering of $X$ formed of sets belonging to $\mathfrak{G}_\alpha$ (resp. $\mathfrak{F}_\alpha$). Now prove that there exists a partition $(B_n)$ of $X$ such that $B_n \subset A_n$ for all $n$ and that $B_n$ belongs to $\mathfrak{F}_\alpha \cap \mathfrak{G}_\alpha$ (use $d$).