Dimension score for metric spaces $\mathbb{R}^n$

Asked Oct 20 '14 at 05:59

Active Oct 25 '14 at 16:21

Viewed 91 times

Given any metric space $(X,d)$, define its score $S(X)$ to be the smallest value of $k$ such that for every $x\in X$ and $r>0$, the ball $B(x,r)$ is covered by at most $2^k$ balls of radius $r/2$.

What is the growth of $S(\mathbb{R}^n)$? I guess it's $\Theta(n)$?

($S(\mathbb{R})=1$, and I'm not sure about $S(\mathbb{R}^2)$.)

edited Oct 25 '14 at 16:21

asked Oct 20 '14 at 05:59

simmons

2,653

$k$ is the smallest number with at most $2^k$ balls , could you elaborate the terminology ? – Souvik Dey Oct 25 '14 at 12:22
@SouvikDey I made a typo, which may have confused you. Just edited, is it clearer? – simmons Oct 25 '14 at 16:22
3

This was asked and answered on MO: Doubling dimension of a Euclidean space. (I think the term "doubling dimension" is in wider usage than "score"). – Oct 25 '14 at 23:47

Dimension score for metric spaces $\mathbb{R}^n$

0 Answers0