Quotients of countable complete metric spaces

Abstract

PROOF. For each y E Y, let Xy be Y with Y { y} made discrete (i.e., Xy is the set Y, topologized by calling open all sets U U S with U open in Y and S C Y { y}). Let X = yE y Xy (the topological sum of the XY), and let f: X -Y be the obvious map.2 The definitions imply thatf is quotient. Clearly X is countable. To show that X is completely metrizable3, it suffices to show that each Xy has this property. Now Xy is a regular space with a countable base, and is therefore metrizable. Moreover, X is the union of the two discrete-hence absolute Gs-subsets { y} and Xvy), so Xy is also an absolute Gs and therefore completely metrizable. That completes the proof. REMARK 1. The conclusion of Theorem 1 remains valid if Y is only assumed to be a countable space with a countable base. (In fact, the proof of [4, Lemma 4.2] implies that such a Y is a continuous open-hence quotient-image of a countable Hausdorff space Y' with countable base, and the proof of Theorem 1 goes through to show that such a Y' is the quotient image of a countable complete metric space.) Our next remark will be applied in [5]. REMARK 2. Theorem 1 can be generalized as follows: Suppose Y is a firstcountable, regular space with a countable, closed subset A for which Y A is completely metrizable. Then Y is a countable-to-one4 quotient image of a complete metric space. PROOF. For each y E A, let Xy be Y with A-{ y) made discrete, and let X = EyEEA XY. Then each Xy is metrizable by the Nagata-Smirnov theorem; the rest of the proof is essentially the same as for Theorem 1.