Equal balls in metric space Let $x$ and $y$ be points in a metric space and let $B(x,r)$ and $B(y,s)$ be usual open balls. Suppose $B(x,r)=B(y,s)$. Must $x=y$? Must $s=r$? 
What I got so far is that: $$r \neq s \implies x \neq y$$ but that's it.
 A: No, it’s not necessary that $x=y$ or that $s=r$. Consider the discrete metric $d$ on a set $X$:
$$d(x,y)=\begin{cases}
0,&\text{if }x=y\\
1,&\text{if }x\ne y\;.
\end{cases}$$
Then $B(x,r)=B(x,s)=\{x\}$ whenever $0<r,s\le 1$, and $B(x,r)=B(y,s)=X$ whenever $r,s>1$.
Added: If $\langle X,d\rangle$ is an ultrametric space, then $B(x,r)=B(y,r)$ whenever $d(x,y)<r$. An example is the product space $\{0,1\}^{\Bbb N}$, where $\{0,1\}$ has the discrete topology, and for $x=\langle x_n:n\in\Bbb N\rangle$ and $y=\langle y_n:n\in\Bbb N\rangle$ we define 
$$d(x,y)=\begin{cases}
0,&\text{if }x=y\\
2^{-m(x,y)},&\text{if }x\ne y\;,
\end{cases}$$
where $m(x,y)=\min\{n\in\Bbb N:x_n\ne y_n\}$. For any finite sequence $\langle x_0,\ldots,x_{n-1}\rangle$ of zeroes and ones, if $y=\langle y_k:k\in\Bbb N\rangle$ and $z=\langle z_k:k\in\Bbb N\rangle$ are such that $y_k=z_k=x_k$ for $k<n$, and $2^{-(n+1)}<r,s\le 2^{-n}$, then $B(y,r)=B(z,s)$.
A: Think minimally: Let $X=\{x,y\}$ be a set with two points. Define $d(x,y)=d(y,x)=1$, and $d(x,x)=0=d(y,y)$.
Then, $B(x,2)=B(y,3)$, yet $x\ne y$ and $2\ne 3$.
A: Let $(X,d)$ be any metric space and let $(X,d')$ be the same set but with the metric $$d'(x,y)=\min\{d(x,y),1\}$$
For all $r,s\geq 1$ and $x,y\in X$, we have $B_{d'}(x,r)=X=B_{d'}(y,s)$.
A: If we consider a discrete metric space (X,d) and let r,s >1 (r not equal to s) and x,y be distinct elements of X. Then B(x,r) = B(y,s).
