= Solution
For $(i)\Rightarrow(ii)$, suppose $B=U\cap F$ with $U$ open and $F$ closed. Since $B\subseteq F$, also $\overline B\subseteq F$. Therefore $U\cap\overline B=B$, showing that $B$ is relatively open in its <closure>.
For $(ii)\Rightarrow(iii)$, write $B=U\cap\overline B$ with $U$ open in $X$. Then $\overline B\setminus B=\overline B\cap(X\setminus U)$ is an intersection of closed sets, hence closed in $X$.
For $(iii)\Rightarrow(i)$, set $F=\overline B$ and $U=X\setminus(\overline B\setminus B)$. The assumed condition makes $U$ open; $F$ is closed and $U\cap F=B$. Thus \b[the three conditions are equivalent]. They characterize a <locally closed subset>.
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