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 $d'(x,y) \le \rho/4 < r$.
Here is a counterexample. Let $X = \mathbb{R}$ and $d_1(x,y) = \lvert y - x \rvert$. Define $d_2(x,y) = \lvert y - x \rvert$ if $\lvert y - x \rvert \le 1$ and $d_2(x,y) = \sqrt{\lvert y - x \rvert}$ if $\lvert y - x \rvert \ge 1$.
An obvious property of $d_2$ is that $\lvert y - x \rvert \le \lvert y' - x' \rvert$ implies $d_2(x,y) \le d_2(x',y')$. We shall moreover need the following well-known fact:
$(*)$ If $1 \le a \le b$, then $\sqrt{b} - \sqrt{a} \le b - a$.
Let us verify that $d_2$ is a metric. The only thing which is not trivial is the triangle inequality. For $x,y,z \in \mathbb{R}$ we have to show $d_2(x,y) \le d_2(x,z) + d_2(z,y)$. This is trivially true if $y = x$. In case $y \ne x$ we may w.l.o.g. assume that $y > x$, i.e. $y = x + r$ with $r > 0$.
(a) $z \le x$. Then $\lvert y - x \rvert \le \lvert z - x \rvert$, hence $d_2(x,y) \le d_2(x,z)$. The triangle inequality follows.
(b) $z \ge y$. Similar!
(c) $x < z < y$.
(c1) $r \le 1$. This reduces to the triangle inequality for $d_1$.
(c2) $r > 1$. Write $z = x + s$ with $0 < s < r$.
(c2.1) $s \le 1$ and $r - s \le 1$. Then we have to show $\sqrt{r} \le s + (r - s) = r$ which is true since $r > 1$.
(c2.2) $s \le 1$ and $r - s > 1$. Then we have to show $\sqrt{r} \le s + \sqrt{r - s}$. This follows from $(*)$.
(c2.3) $s > 1$ and $r - s \le 1$. Then we have to show $\sqrt{r} \le \sqrt{s} + (r - s) $. This follows agaim from $(*)$.
(c2.4) $s > 1$ and $r - s > 1$. Then we have to show $\sqrt{r} \le \sqrt{s} + \sqrt{r - s}$. This is obvious (take the squares of both sides).
$d_1$ are $d_2$ are uniformly equivalent because $d_2(x,y) \le d_1(x,y)$ and $d_2(x,y) < \min(1,\epsilon)$ implies $d_1(x,y) < \epsilon$ (in fact, we have $d_2(x,y) < 1$ which is only possible when $\lvert y - x \rvert < 1$ so that $d_1(x,y) = d_2(x,y) < \epsilon $).
$d_1$ and $d_2$ have the same bounded sets. If $B$ is bounded with respect to $d_1$, then it is obviously respect to $d_2$. Let $B$ be unbounded with respect to $d_1$. Then there exist $x_n,y_n \in B$ such that $d_1(x_n,y_n) \ge n$. But then $d_2(x_n,y_n) \ge \sqrt{n}$, hence $B$ be unbounded with respect to $d_2$.
$d_1$ and $d_2$ are not Hölder equivalent. Assume there are $C > 0$ and $\alpha \in (0,1]$ such that $d_1(x,y) \le C (d_2(x,y))^\alpha$. For $\lvert y - x \rvert \ge 1$ this means $\lvert y - x \rvert \le C \lvert y - x \rvert^{\alpha/2}$, i.e. $\lvert y - x \rvert^{1 - \alpha/2} \le C$. But $1 - \alpha/2 \in [1/2,1)$, hence $\lvert y - x \rvert^{1 - \alpha/2}$ cannot bounded by any constant $C$. This is a contradiction.
Best Answer
The textbook you linked distinguishes between $3$ types of equivalence:
•strong equivalence
•uniform equivalence &
•equivalence.
Evidently the condition that the identity map and its inverse be uniformly continuous corresponds to uniform equivalence. But strong equivalence is stronger, and is the one for which bounded sets are preserved.
As pointed out in the text, $\rho\le C\sigma$ may fail under uniform equivalence. In particular, it may fail when $\sigma =\frac{\rho}{1+\rho}$.