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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?

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  • $\begingroup$ Do you allow $r = \infty$? And what is the box dimension? $\endgroup$ – Paul Frost Dec 25 '18 at 11:31
  • $\begingroup$ @PaulFrost No, infinity is not allowed. Here is what box dimension is: en.wikipedia.org/wiki/Minkowski–Bouligand_dimension In any case, I found the answer to that part of my question here: math.stackexchange.com/a/964025/71829 $\endgroup$ – Keshav Srinivasan Dec 25 '18 at 15:09
  • $\begingroup$ So you only consider totally bounded metric spaces. $\endgroup$ – Paul Frost Dec 25 '18 at 15:13
  • $\begingroup$ @PaulFrost What gave you that idea? $\endgroup$ – Keshav Srinivasan Dec 25 '18 at 15:13
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    $\begingroup$ Two of my comments are nonsense. What I wanted to ask is this. If $n(r)$ is the $r$-covering number of $X$, do you allow $n(r) = \infty$? Only if you do not, my comment concerning totally bounded metric spaces makes sense. $\endgroup$ – Paul Frost Dec 26 '18 at 9:06
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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 $d'(x,y) \le \rho/4 < r$.

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