Can a metric be recovered from the collection of open balls it produces? My book says that this cannot be done unless the radius and the centre of the ball are known.
I don't understand, why is it important to know  the radius and the centre of the ball ?
 A: Here's a simple example: Consider the two metrics $d(x,y) = |x-y|$ and $d'(x,y) = 2|x-y|$ on $\mathbb{R}$. They both produce the same balls but with different radii: $\{x \in\mathbb{R} : d(x,x_0) < r\} = \{x\in\mathbb{R} : d'(x,x_0) < r/2\}$.
A: A nice example, with a non-archimedean metric: Consider the rational $\mathbb Q$, with the metric $d(x,y)=|x-y|_2$, where $|\cdot|_2$ is defined as follows: $|0|_2=0$ and, given a rational $r\ne 0$, write it as $\displaystyle r=2^a\frac bc$, where $a$ is an integer (maybe $0$, maybe negative) and $b,c$ are odd integers, and set $|r|_2=2^{-a}$. This is the $2$-adic distance, with $|\cdot|_2$ being the $2$-adic norm.
The triangle inequality follows from the stronger condition $|x+y|_2\le\max\{|x|_2,|y|_2\}$, which implies that every triangle is isosceles. From this, it follows that every ball has many centers. For example, if $s\in B_1(r)$, the ball of radius $1$ around $r$, then in fact $B_1(s)=B_1(r)$, that is, any point in the ball is a center of the ball. 
In this example, we only have one metric, but given a family of balls, there is no way for us to know which point in the balls was "intended" as their center, unless we are told. 
(Moreover, the distance $d'(x,y)=|x-y|'$ where $\displaystyle\left|2^a\frac bc\right|'=\pi^{-a}$ has the same balls, so even if we are told the intended centers, we still cannot recover the distance.)
A: Let $X=\{a,b\}$, where $a\ne b$. Define metrics $d_1$ and $d_2$ on $X$ as follows:
$$d_1(x,y)=\begin{cases}0,&\text{if }x=y\\1,&\text{if }x\ne y\end{cases}$$
and
$$d_2(x,y)=\begin{cases}0,&\text{if }x=y\\2,&\text{if }x\ne y\end{cases}\;.$$
Check that in both $\langle X,d_1\rangle$ and $\langle X,d_2\rangle$ the open balls are the sets $\{a\}$, $\{b\}$, and $X$, even though the metrics are clearly different, since $d_1(a,b)=1\ne2=d_2(a,b)$.
