Totally bounded metric space

Question

Consider $\Bbb{R}$ with the following metric: $$\rho(x,y):=\left|\frac{x}{1+|x|}-\frac{y}{1+|y |}\right|$$ Then I wish to show that $(\Bbb{R}, \rho)$ is a totally bounded metric space.

Of course by totally bounded I mean that $\forall\epsilon>0$ there exists a finite set $S\subset\Bbb{R}$ such that $\Bbb{R}\subset\bigcup_{x \in S} B_{\epsilon}(x)$.

Now this metric space is not complete, since $x_n=n$ is a Cauchy sequence in $(\Bbb{R}, \rho)$ which does not converge in the space, so I cannot use compactness. I have tried quite a few different things, all to varying degrees of success. How would I go about proving this?

You can consider the completion and show that that is compact. — Daniel Fischer, Oct 30 '13 at 23:12

score 1 · Accepted Answer · answered Oct 30 '13 at 23:13

1

HINT: Show that the map

$$h:\Bbb R\to(-1,1):x\mapsto\frac{x}{1+|x|}$$

is an isometry from $\langle\Bbb R,\rho\rangle$ to $\langle(-1,1),d\rangle$, where $d$ is the usual metric on $(-1,1)$; it’s easy to show that $\langle(-1,1),d\rangle$ is totally bounded.

answered Oct 30 '13 at 23:13

Brian M. Scott

616,228

Shall we duel? ${}{}{}$ – Pedro Oct 30 '13 at 23:17
Ah, thank you kindly. I'm not sure why I overlooked constructing an isometry - always seems clear when someone else has pointed it out. – George Oct 30 '13 at 23:19
@George: You’re very welcome. – Brian M. Scott Oct 30 '13 at 23:19
@Pedro: I win: $8>6$, so $h$ beats $f$. :-) – Brian M. Scott Oct 30 '13 at 23:20
@BrianM.Scott Darn! ${}{}{}{}$ – Pedro Oct 30 '13 at 23:22

score 1 · Answer 2 · answered Oct 30 '13 at 23:14

1

Let $f:\Bbb R\to (-1,1)$ with $f(x)=\dfrac{x}{1+|x|}$. Then $f$ is an isometry of $(\Bbb R,\rho)$ with $((-1,1),d)$ where $d$ is the usual Euclidean distance.

answered Oct 30 '13 at 23:14

Pedro

122,002

Totally bounded metric space

2 Answers2