If $x\ne b$, then $d_a(b,x)=d(b,a)+d(a,x)\ge d(b,a)$. The strict open-ball inequality therefore excludes every such $x$, while including $b$. Hence $\boxed{B_{d_a}(b,d(b,a))=\{b\}}$. Every point other than $a$ is isolated in this metric.


Solution

Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / E / E3 / b / ii / Solution

 Solution (source code)