Let $(X,d)$ be a metric space and assume that $B_r^d(x)=B_s^d(y)$. is $r=s$ and $x=y$

Question

Let $(X,d)$ be a metric space and assume that $B_r^d(x)=B_s^d(y)$ where: $$B_r^d=\{ a \in X | d(a,x) < r\}$$ Now, is it always true that

(a) $r=s$

(b) $x=y$

I made an elaborate argument on this question why both these statements should be true. However, I later doubted this conclusion. Lets take the following metric: $$d(x,y)=1 \text{ if } x\neq y,\ 0 \text{ if } x=y$$ Then $B_{0.5}^d(x)=B_{0.4}^d(x)=\{x\}$ assuming for the moment that the balls have the same centre. Does this mean that $r$ does not necessarily equal $s$? And how about if they do not have the same centre?

Answer 1

Using your example $d$ (called the discrete metric), note that if $r > s > 1$ and $x \neq y$, then $B_r^d(x) = B_s^d(y) = X$ (the entire space). So both statements are not necessarily true.

Answer 2

This is not true. For example in an ultrametric space you have $d(x,y) < r \implies B_r(x) = B_r(y)$, i.e. any point of a ball is its centre.