{c}

We prove (ii)$\Rightarrow$(iii). If $B$ is open in $\overline B$, its complement $\overline B\setminus B$ is closed in the subspace $\overline B$. A relatively closed subset of a closed subset of $X$ is closed in $X$: it is the intersection of that closed subset with another closed subset of $X$. Since a <closure> is closed, \b[$\overline B\setminus B$ is closed in $X$].

Solution

Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / E / 1 / b / ii / Solution

 Solution (source code)