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I vaguely remember a theorem that says that any two metrics on the Euclidean space $\mathbb{R}^n$ are equivalent in some sense, but probably not in the sense of metric equivalence: two metrics $d_1$ and $d_2$ are said to be metrically equivalent if there are positive numbers $c_1$ and $c_2$ such that

$c_1 d_1(x,y) \le d_2(x,y) \le c_2 d_1(x,y)$, $\forall$ $x,y \in \mathbb{R}^n$

Can somebody confirm which kind of equivalence there might be?

Thanks.

Michael C
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  • I personally don't know of any definition of equivelance between metrics other than the one you stated. – Git Gud Jan 16 '13 at 22:41
  • @GitGud: There are many. For example, you might consider two metrics equivalent if they induce the same topology, or the same uniform structure. – Chris Eagle Jan 16 '13 at 22:42
  • @GitGud Here's another: we call the metrics $d_1$ and $d_2$ on the same space $X$ equivalent if for every $x \in X$ and $r > 0$ there exist $r_1,r_2 > 0$ such that $N_{r_1}^{(1)}(x) \subset N_r^{(2)}(x)$ and $N_{r_2}^{(2)}(x) \subset N_r^{(1)}(x)$, where $N_\alpha^{(i)}(x)$ is the $\alpha$-ball in the $d_i$ metric. – Gyu Eun Lee Jan 16 '13 at 22:45
  • @proximal: That's the same thing as inducing the same topology. – Chris Eagle Jan 16 '13 at 23:02
  • @ChrisEagle Ah. Indeed it is. (I learned this definition in real analysis class, so I hadn't thought about it in terms of inducing topologies.) – Gyu Eun Lee Jan 16 '13 at 23:14

1 Answers1

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The correct statement is that any two metrics derived from norms on $\Bbb{R}^n$ are equivalent in this sense. There are plenty of metrics that are not equivalent: for example, the discrete metric is not equivalent to the standard Euclidean metric, but this is because the discrete metric doesn't come from a norm.

Chris Eagle
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