= Solution
Put $A=\mathcal O(X)$ and $\neg a=a\to0$. The closure of the open $a$ is the <closed sublocale> with <closed nucleus> $c(\neg a)$; its interior is the open $\neg\neg a$. The closed <sublocale> complementary to $\neg a$ is also open precisely when $\neg a$ is complemented, in which case its open representative is $\neg\neg a$. Thus the closure is clopen precisely when the <regular open element> $\neg\neg a$ is complemented. Every <regular open element> has this form, so \b[(i) and (ii) are equivalent].
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