Let $d_1(x,y)$ and $d_2(x,y)$ be any two metrics on $\mathbb{R}^n$. Can it be shown that,
$$c\cdot d_2(x,y) \le d_1(x,y) \le C\cdot d_2(x,y)$$
for all $x,y \in \mathbb{R}^n$ for some fixed positive constants $c,C$? If not, under what conditions could such a relation hold (for a compact set it seems straightforward)? If yes, does this result hold for two arbitrary topological metric spaces as well?
Thanks!
It's obviously not true if you really mean to ask about any two metrics; as people have pointed out you could let $d_1$ be the standard metric and $d_2$ the discrete metric.
A less silly version of the question, perhaps what you actually meant to ask, in any case what I assumed you meant when I read the question, is this: Suppose $d_1$ and $d_2$ are metrics on $\Bbb R^n$, both of which induce the standard topology on $\Bbb R^n$. Does that inequality follow?
The answer to the revised question is still no. Let $d_1(x,y)=|x-y|$ and $d_2(x,y)=\min(1,|x-y|)$.
(It's easy to show that $d_2$ is a metric and that the inequality is false. To show the two metrics induce the same topology you need to show that a set is $d_1$-open if and only if it is $d_2$-open. Hint for that: If $0<r<1$ then $B_{d_1}(x,r)=B_{d_2}(x,r)$.)
This is most definitely not true, as this condition implies that the two metrics induce the same topology. The discrete topology correspond to the metric:
$\rho(x,y)= \begin{cases} 1, & x\neq y \\ 0, & x=y \end{cases}$
The appropriate topological space in this case is the discrete topology (every subset is open) which is not connected, but the euclidean metric does induce a connected topological space.
If $d_1$ is the usual metric and $d_2$ is the discrete metric, then there are no such constants.
However, your statement is true if the metrics are induced by norms in $\mathbb{R}^n$.
