no countable dense subset of a complete metric space is a
Let be a complete metric space with no isolated points, and let be a countable dense set. Then is not a set (http://planetmath.org/G_deltaSet).
Proof
First, we will prove that is first category. Then, supposing that is a , we will conclude that must be first category. But then so must be , which is absurd because is complete.
1) is a first category set:
By hypothesis . Let’s see that each singleton is nowhere dense if has no isolated points:
(trivially). Suppose that . Then there is a ball aisolating the point. Absurd ( has no isolated points). Then and we have that so every singleton is nowhere dense and is of first category because it is a countable union of nowhere dense sets.
2) Suppose is a , that is, such that every is open. As is dense, then each is dense, because . But then and
which implies that is of first category. Then is of first category. Absurd, because is complete.
Title | no countable dense subset of a complete metric space is a |
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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 |