 Solution (source code)

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= Solution
{c}

We prove (iii)$\Rightarrow$(i). If $C=\overline B\setminus B$ is closed in $X$, set $U=X\setminus C$ and $F=\overline B$. Then $U$ is open, $F$ is closed, and
$$
\boxed{U\cap F=(X\setminus(\overline B\setminus B))\cap\overline B=B.}
$$
Together the three implications prove the equivalence. These are the standard characterizations of a <locally closed subset>.

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