Can any metric on $\Bbb R^n$ be bounded above and below for any other metric? 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!
 A: 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 $B_{d_1}(x,r)=B_{d_2}(x,r)$.)
A: 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. 
A: 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 $\mathbb{R}^n$.
