Exercise 4, § 6

Let $X$ be a metrizable space.

a) Show that the set of Borel subsets of $X$ is the smallest subset $\mathfrak{F}$ of $\mathscr{P}(X)$ which contains all closed subsets of $X$ and is such that countable unions and countable intersections of sets of $\mathfrak{F}$ belong to $\mathfrak{F}$. (Consider the subset $\mathfrak{S}$ of $\mathfrak{F}$ consisting of all $A \in \mathfrak{F}$ such that $X - A \in \mathfrak{F}$.)

b) Show that the set of Borel subsets of $X$ is the smallest subset $\mathfrak{F}$ of $\mathscr{P}(X)$ which contains all open subsets of $X$ and is such that every countable intersection of sets of $\mathfrak{F}$ belongs to $\mathfrak{F}$, and every union of a sequence of mutually disjoint sets of $\mathfrak{F}$ belongs to $\mathfrak{F}$ (same method).

c) Every ordinal $\alpha$ can be written uniquely in the form $\alpha = \omega \beta + n$, where $n < \omega$ (Set Theory, Chapter III, § 2, Exercise 15); $\alpha$ is said to be even or odd according as $n$ is even or odd. For every countable ordinal $\alpha$, we define by transfinite induction 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 all closed (resp. open) subsets of $X$; if $\alpha$ is even and $> 0$, $\mathfrak{F}_{\alpha}$ (resp. $\mathfrak{G}_{\alpha}$) is the set of all subsets of $X$ which are countable intersections (resp. countable unions) of sets belonging to $\bigcup_{\xi < \alpha} \mathfrak{F}_{\xi}$ (resp. $\bigcup_{\xi < \alpha} \mathfrak{G}_{\xi}$); if $\alpha$ is odd, then $\mathfrak{F}_{\alpha}$ (resp. $\mathfrak{G}_{\alpha}$) is the set of all subsets of $X$ which are countable unions (resp. countable intersections) of sets belonging to $\bigcup_{\xi < \alpha} \mathfrak{F}_{\xi}$ (resp. $\bigcup_{\xi < \alpha} \mathfrak{G}_{\xi}$). Show that the union of the $\mathfrak{F}_{\alpha}$ (resp. $\mathfrak{G}_{\alpha}$) 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 power at most equal to the power of the continuum.

d) The relation $A \in \mathfrak{F}_{\alpha}$ is equivalent to $\complement A \in \mathfrak{G}_{\alpha}$. 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 $\mathfrak{G}_{\alpha} \subset \mathfrak{F}_{\alpha+1} \cap \mathfrak{G}_{\alpha+1}$ (use transfinite induction).

e) Let $f$ be a continuous mapping of $X$ into a metrizable space $Y$. Show that $\overline{f}^{-1}(\mathfrak{F}_\alpha(Y)) \subset \mathfrak{F}_\alpha(X)$ and $\overline{f}^{-1}(\mathscr{G}_\alpha(Y)) \subset \mathscr{G}_\alpha(X)$.

f) If $X$ and $Y$ are two metrizable spaces and $A \in \mathfrak{F}_\alpha(X)$, $B \in \mathfrak{F}_\alpha(Y)$, [resp. $A \in \mathscr{G}_\alpha(X)$, $B \in \mathscr{G}_\alpha(Y)$], then $A \times B \in \mathfrak{F}_\alpha(X \times Y)$ [resp. $A \times B \in \mathscr{G}_\alpha(X \times Y)$].

g) Let $(X_n)$ be a sequence of metrizable spaces. Show that if $A_n$ is a Borel set in $X_n$ for each $n$, then $\prod_n A_n$ is a Borel set in $\prod_n X_n$.

h) Let $\alpha > 0$ be an even (resp. odd) countable ordinal and let $(A_n)$ be a countable covering of $X$ formed of sets belonging to $\mathscr{G}_\alpha$ (resp. $\mathfrak{F}_\alpha$). Show that there is a partition $(B_n)$ of $X$ such that $B_n \subset A_n$ and $B_n \in \mathfrak{F}_\alpha \cap \mathscr{G}_\alpha$ for each $n$ [use d)].