Exercise 7, § 6

a) Let $J$ be the Polish space $\mathbf{N}^\mathbf{N}$ (where $\mathbf{N}$ is endowed with the discrete topology), $X$ a metrizable space of countable type. Now prove that for every countable ordinal $\alpha$, there exists a subset $G_\alpha \subset J \times X$ such that: $1^o$ $G_\alpha \in \mathfrak{G}_\alpha(J \times X)$ (Exercise 6); $2^o$ for every subset $U \in \mathfrak{G}_\alpha(X)$, there exists $z \in J$ such that $G_\alpha(z) = U$. One can 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 every integer $n$, let $f_n = \mathrm{pr}_n \circ f$. If $(A_n)$ is a countable basis of $X$, containing the empty set, one takes $G_0$ such that $G_0(z) = \bigcup_n A_{f_n(z)}$ for every $z \in J$. For every countable ordinal $\alpha$, one takes $G_{\alpha+1}$ such that $G_{\alpha+1}(z) = \bigcap_n G_\alpha(f_n(z))$ if $\alpha$ is pair, $G_{\alpha+1}(z) = \bigcup_n G_\alpha(f_n(z))$ if $\alpha$ is odd. Finally, if $\alpha$ has no predecessor and if $(\lambda_n)$ is a increasing sequence of ordinals such that $\alpha = \sup_n \lambda_n$, one takes $G_\alpha$ such that $G_\alpha(z) = \bigcup_n G_{\lambda_n}(f_n(z))$. One will use the fact that $f_n$ is continuous in $J$.

b) One takes in particular $X = J$. Now prove that the set $\mathrm{pr}_1(G_\alpha \cap \Delta)$, where $\Delta$ is the diagonal of $J \times J$, belongs to $\mathfrak{G}_\alpha(J)$, but not to $\mathfrak{F}_\alpha(J)$ (reason by contradiction using $a$).