Proposition 9

Let X be a metrizable space, and let A be a relatively compact Souslin subspace of X. Then there exists a compact metrizable space K, a decreasing sequence $(B_n)$ of subsets of K, each of which is a countable union of compact sets, and a continuous mapping $f : K \to X$, such that $A = f\left(\bigcap_n B_n\right)$.

Replacing X by $\overline{A}$ if necessary, we may suppose that X is compact and that A is dense in X. Since A is a Souslin space, there is a Polish space P and a continuous mapping $g : P \to X$ such that $g(P) = A$. By no. 1, Theorem 1, Corollary 1, we may assume that P is the intersection of a decreasing sequence $(U_n)$ of open subsets of the cube $I^\mathbf{N}$. Let K be the space $I^\mathbf{N} \times X$, which is compact and metrizable (§ 2, no. 4, Theorem 1, Corollary 2). Let $G \subset P \times X$ be the graph of g, let $\overline{G}$ be the closure of G in K, and let f denote the projection of $K = I^\mathbf{N} \times X$ onto X; then clearly we have $f(G) = A$. Since g is continuous, G is closed in $P \times X$ (Chapter I, § 8, no. 1, Proposition 2, Corollary 2) and $G = \overline{G} \cap (P \times X)$; hence $G = \bigcap_n B_n$, where

Since each $U_n$ is a countable union of closed sets in $I^\mathbf{N}$ (§ 2, no. 5, Proposition 7), each $B_n$ is a countable union of compact sets and the proof is complete.