Let $(X,d)$ be a metric space. Then for any $r>0$, the $r$-covering number of $X$ is the minimum number of open balls of radius $r$ needed to cover $X$. And if $d_1$ and $d_2$ be two metrics on the same set $X$, then $d_1$ and $d_2$ are called strongly equivalent if there exist constants $\alpha,\beta>0$ such that $\alpha d_1(x,y)\leq d_2(x,y)\leq\beta d_1(x,y)$ for all $x,y\in X$.
My question is, if two metrics are strongly equivalent, then do they have the same $r$-covering number for all $r>0$? If not, do they at least have the same box dimension?
Best Answer
Let $X = [0,2]$ with the usual metric $d(x,y) = \lvert x - y \rvert$. Then the $1$-covering number is $2$.
Define $d'(x,y) = \min(1/2,d(x,y))$. This is a metric such that $$d'(x,y) \le d(x,y) \le 4 d'(x,y) .$$ (If $d(x,y) \le 1/2$, then $d(x,y) = d'(x,y)$, otherwise $d(x,y) \le 2 = 4 \cdot 1/2 = 4 d'(x,y)$.)
Its $1$-covering number is $1$.
Edited:
Here is a more general answer.
Let $n(r) = n(r,d)$ denote the $r$-covering number of $X$. If the metric $d$ is unbounded (i.e. $diam(X,d) = \sup_{x,y \in X} d(x,y) = \infty$), then all $n(r) = \infty$.
Let $d,d'$ be strongly equivalent.
(1) $d$ is bounded if and only idf $d'$ is bounded.
This is obvious.
(2) If $d$ is unbounded, then $n(r,d) = n(r,d') = \infty$ for all $r$.
This is again obviuos.
(3) If $d$ is bounded, then there exists a metric $d'$ strongly equivalent to $d$ and $r$ such that $n(r,d) > n(r,d') = 1$.
Let $\rho = diam(X,d)$. Define $d'(x,y) = \min(\rho /4, d(x,y))$. This is a metric such that $$d'(x,y) \le d(x,y) \le 4 d'(x,y) .$$ (If $d(x,y) \le \rho /4$, then $d(x,y) = d'(x,y)$, otherwise $d(x,y) \le \rho = 4 \cdot \rho /4 = 4 d'(x,y)$.)
Let $\rho/4 < r < \rho/2$.
Then $n(r,d) > 1$ because a single open ball of radius $r$ cannot cover $X$ (Assume $B(x,r) = X$ for some $x \in X$. Then for all $y,z \in X$ one has $d(y,z) \le d(y,x) + d(y,z) < 2r$, hence $diam(X,d) \le 2r < \rho$.)
On the other hand $n(r,d') = 1$. For any $x \in X$ we have $B(r,d') = X$ because for any $y \in Y$ we have $d'(x,y) \le \rho/4 < r$.