Solution
= Solution
\b[True.] Part (v) supplies one isolated point $p$. Since $\{p\}$ is open, $X\setminus\{p\}$ is closed and therefore complete. It is nonempty when $X$ has at least two points, so part (v) applied to this subspace supplies an isolated point $q$. Because $d(p,q)>0$, a sufficiently small ball showing that $q$ is isolated in the subspace also avoids $p$; hence $q$ is isolated in $X$.