Exercise 11, § 6

(a) Let X be a Polish space, R an equivalence relation in X. Now prove that there exists in X a set which is a countable intersection of open sets, meeting each equivalence class according to R in one and only one point, when one of the following two hypotheses is verified: $\alpha$ R is closed (Follow the method of proof of th. 4 of IX, p. 69, using the following remark: if $(G_n)$ is a sequence of sets each of which is a countable intersection of open sets, and if for every $n$ there exists a neighbourhood $V_n$ of $G_n$ such that the family $(V_n)$ is locally finite, then $\bigcup_n G_n$ is a countable intersection of open sets). $\beta$ X is locally compact and the graph of R is closed in $X \times X$. (Same method, observing that the saturation of a compact set is then closed; cf. I, p. 113, exerc. 16). (b) Let X be a Polish space, R an equivalence relation in X, both open and closed, $\varphi$ the canonical mapping of X onto X/R. Now prove that there exists in X a subset A, a countable intersection of open sets in X, such that the restriction of $\varphi$ to A is a homeomorphism of A onto $\varphi(A)$ and that $\varphi(A)$ is everywhere dense in X/R and a countable intersection of open sets. (With the notations of the proof of th. 4 of IX, p. 69, show that one can suppose for each $n$ the $\varphi_n(c)$ defined in such a way that the union of the interiors of the $h_n(c)$ ($c \in C_n$) has an image in X/R which is everywhere dense; use a) and th. 1 of IX, p. 58.) (c) Now prove that the conclusion of b) is still valid when X is a locally compact space having a countable basis, R a closed equivalence relation in X, such that every class according to R is compact. (Reduce to the case b) by using exerc. 26 of IX, p. 116.)