Line segments in metric space

Question

Is it true in a metric space that $d(x,y)=d(B(x,r),y)+r$, where $B(x,r),$ is a ball centered at $x$ and having radus $r$, and d is distance function?

Answer 1

The statement is generally false. Let, for eample, $d(x,y)=|x-y|, x,y\in \mathbb{R}$ be a metric. Define $d'(x,y)$ by $$d'(x,y)=\mathrm{min}\{ d(x,y),1\}.$$ Indeed, $d'$ is a metric on $\mathbb{R}$. As well, $d'(0,2)=1$. Consider the ball $B_{d'}(0,1/2)$. Clearly, $d'(B_{d'}(0,1/2), 2)=1\neq1+1/2$, so that the statement is false.

Answer 2

If $y \in B(x,r)$ then you want $d(x,y)=r$. In other words, $d(x,y)<r$ implies $d(x,y)=r$. Not true except in trivial cases.