There isn't a complete metric over $\mathbb{Q}$ equivalent to the usual metric.

Question

I thought about proving this proposition showing that $\mathbb{Q}$ isn't homeomorphic to any Baire spaces, but I'm not sure if this is enough (or even the correct path). Could someone help?

look at https://math.stackexchange.com/questions/15710/why-is-it-that-mathbbq-cannot-be-homeomorphic-to-any-complete-metric-spac — Chinnapparaj R, Nov 26 '19 at 00:08
@Timkinsella https://en.wikipedia.org/wiki/Equivalence_of_metrics I'm talking about the topological equivalence — russellsparadox, Nov 26 '19 at 00:13

score 2 · Answer 1 · answered Nov 26 '19 at 00:08

2

That is enough. $\mathbb Q$ is a countable union of nowhere dense sets (namely singletons) so it cannot be complete under any equivalent metric.

answered Nov 26 '19 at 00:08

Kavi Rama Murthy

311,013

Thank you, Kabo Murphy! – russellsparadox Nov 26 '19 at 00:12

There isn't a complete metric over $\mathbb{Q}$ equivalent to the usual metric.

1 Answers1