Exercise 5, § 6

a) Let $J$ be the Polish space $\mathbf{N}^\mathbf{N}$ ($\mathbf{N}$ carrying the discrete topology) and let $X$ be a metrizable space of countable type. Show that for each countable ordinal $\alpha$ there is a subset $G_\alpha$ of $J \times X$ such that (i) $G_\alpha \in \mathscr{G}_\alpha(J \times X)$ (Exercise 4) and (ii) for every subset $U \in \mathscr{G}_\alpha(X)$ there exists $z \in J$ such that $G_\alpha(z) = U$. [We may proceed as follows by transfinite induction: the spaces $J$ and $J^\mathbf{N}$ being homeomorphic, let $f$ be a homeomorphism of $J$ onto $J^\mathbf{N}$, and for each integer $n$ let $f_n = \mathrm{pr}_n \circ f$. If $(A_n)$ is a countable base of $X$ containing the empty set, take $G_0$ such that $G_0(z) = \bigcup_n A_{f_n(z)}$ for each $z \in J$. For each countable ordinal $\alpha$, take $G_{\alpha+1}$ to be such that $G_{\alpha+1}(z) = \bigcap_n G_\alpha(f_n(z))$ if $\alpha$ is even, and $G_{\alpha+1}(z) = \bigcup_n G_\alpha(f_n(z))$ if $\alpha$ is odd. Finally, if $\alpha$ has no predecessor and if $(\lambda_n)$ is an increasing sequence of ordinals such that $\alpha = \sup \lambda_n$, take $G_\alpha$ to be such that $G_\alpha(z) = \bigcup_n G_{\lambda_n}(f_n(z))$. Use the fact that $f_n$ is continuous on $J$.]

b) In particular take $X = J$. Show that the set $\mathrm{pr}_1(G_\alpha \cap \Delta)$, where $\Delta$ is the diagonal of $J \times J$, belongs to $\mathscr{G}_\alpha(J)$ but not to $\mathfrak{F}_\alpha(J)$ [argue by contradiction, using a)].