Can a metric be recovered from the collection of open balls it produces?

Question

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 ?

Answer 1

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: $\{x \in\mathbb{R} : d(x,x_0) < r\} = \{x\in\mathbb{R} : d'(x,x_0) < r/2\}$.

Answer 2

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.)

Answer 3

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 $d_1(a,b)=1\ne2=d_2(a,b)$.