Bounded and unbounded homeomorphic metric spaces

Question

I'm struggling to come up with an example of homeomorphic metric spaces such that one is bounded and one is not

What about $]0;1[$ and $]-\infty;0[$ with $d(x,y) = |x-y|$? – nombre Oct 14 '15 at 18:09 — nombre, Oct 14 '15 at 18:09

score 2 · Answer 1 · answered Oct 16 '15 at 16:13

2

$$ \tan : (-\pi,\pi) \to \mathbb R $$ That is a homeomorphism from $(-\pi,\pi)$ to $\mathbb R$.

So is this: $x \mapsto \dfrac x {\pi^2-x^2}$.

answered Oct 16 '15 at 16:13

Michael Hardy

1

hardmath · Answer 2 · 2015-10-16T16:08:20.193

1

Consider $\mathbb{Z}$ with the discrete topology. Let $d_1(x,x) =0$ and otherwise $1$. Let $d_2 (x,y)=|x-y|$. Both metrics produce the discrete topology, and only one of them is bounded.

edited Oct 16 '15 at 16:08

answered Oct 14 '15 at 20:49

hardmath

37,015

score 0 · Answer 3 · answered Oct 14 '15 at 18:25

If you consider $\mathbb{R}$ and $]-\frac{\pi}{2},\frac{\pi}{2}[$ as metrics spaces for the usual metric, then you can check that $\arctan:\mathbb{R}\to]-\frac{\pi}{2},\frac{\pi}{2}[$ is an homeomorphism. For this metric, you have that $\mathbb{R}$ is not bounded whereas $]-\frac{\pi}{2},\frac{\pi}{2}[\subset B(0,2)$.

Bounded and unbounded homeomorphic metric spaces

3 Answers3