no countable dense subset of a complete metric space is a Gδ

Let (X,d) be a complete metric space with no isolated pointsMathworldPlanetmath, and let DX be a countableMathworldPlanetmath dense set. Then D is not a Gδ set (


First, we will prove that D is first category. Then, supposing that D is a Gδ, we will conclude that X-D must be first category. But then so must be X, which is absurd because X is completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath.

1) D is a first category set:

By hypothesisMathworldPlanetmathPlanetmath D={xi}iN. Let’s see that each singleton is nowhere dense if X has no isolated points:

{xi}¯={xi} (trivially). Suppose that {xi}o={xi}. Then there is a ball aisolating the point. Absurd (X has no isolated points). Then {xi}o= and we have that ({xi}¯)o= so every singleton is nowhere dense and D is of first category because it is a countable union of nowhere dense sets.

2) Suppose D is a Gδ, that is, D=i=1Ui such that every U is open. As D is dense, then each U is dense, because DUiD¯=XUi¯i. But then X-D=X-i=1Ui=i=1(X-Ui) and (X-Ui¯)o=(X-Uio)o=(X-Ui)o=X-Ui¯=

which implies that X-D is of first category. Then D(X-D)=X is of first category. Absurd, because X is complete.

Title no countable dense subset of a complete metric space is a Gδ
Canonical name NoCountableDenseSubsetOfACompleteMetricSpaceIsAGdelta
Date of creation 2013-03-22 14:59:06
Last modified on 2013-03-22 14:59:06
Owner gumau (3545)
Last modified by gumau (3545)
Numerical id 8
Author gumau (3545)
Entry type Result
Classification msc 54E52