# Are equivalent metrics comparable on compact sets?

• Aug 17th 2010, 03:38 AM
yoyoma
Are equivalent metrics comparable on compact sets?
Suppose I have a set $\displaystyle X$ equipped with two equivalent metrics, $\displaystyle d_1$ and $\displaystyle d_2$, meaning that the metrics induce the same topology on $\displaystyle X$. I know that $\displaystyle d_1$ and $\displaystyle d_2$ need not be comparable on all of $\displaystyle X$, however, is it true that they will be comparable on compact subsets of $\displaystyle X$?
• Aug 18th 2010, 06:35 AM
Iondor
No, they aren't.
Consider the cantor space $\displaystyle C=\{0,1\}^{\mathbb{N}}$ , which is the product space of the discrete space {0,1}.
It is compact and induced by two incomparable metrices
$\displaystyle d_{1}(x,y)=2^{-n}$
$\displaystyle d_{2}(x,y)=3^{-n}$
, where $\displaystyle n\in \mathbb{N}$ is the smallest i , such that $\displaystyle x_{i} \neq y_{i}$

Then there is no constant K such that
$\displaystyle d_{1}(x,y)< K\cdot d_{2}(x,y)$ for all x,y.
• Aug 18th 2010, 07:23 AM
yoyoma
Thanks for your reply. May I also ask, what about in the situation where the distance functions $\displaystyle d_1$ and $\displaystyle d_2$ are induced by a Riemannian metric on a smooth manifold? In this case the topologies induced by $\displaystyle d_1$ and $\displaystyle d_2$ 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?