Are equivalent metrics comparable on compact sets?

**yoyoma** · August 17th 2010, 03:38 AM

Suppose I have a set $X$ equipped with two equivalent metrics, $d_1$ and $d_2$ , meaning that the metrics induce the same topology on $X$ . I know that $d_1$ and $d_2$ need not be comparable on all of $X$ , however, is it true that they will be comparable on compact subsets of $X$ ?

**Iondor** · August 18th 2010, 06:35 AM

No, they aren't.
Consider the cantor space $C=\{0,1\}^{\mathbb{N}}$ , which is the product space of the discrete space {0,1}.
It is compact and induced by two incomparable metrices
$d_{1}(x,y)=2^{-n}$
$d_{2}(x,y)=3^{-n}$
, where $n\in \mathbb{N}$ is the smallest i , such that $x_{i} \neq y_{i}$

Then there is no constant K such that
$d_{1}(x,y)< K\cdot d_{2}(x,y)$ for all x,y.

**yoyoma** · August 18th 2010, 07:23 AM

Thanks for your reply. May I also ask, what about in the situation where the distance functions $d_1$ and $d_2$ are induced by a Riemannian metric on a smooth manifold? In this case the topologies induced by $d_1$ and $d_2$ each agree with the manifold topology, but these distance functions need not be comparable on the entire manifold. However, will they be comparable on compact subsets of the manifold?

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