 Solution

Suppose every regular open element is complemented. In a Heyting algebra, a complemented element is regular: if $a\vee b=1$ and $a\wedge b=0$, then $b=\neg a$ and $a=\neg b$. Finite joins of complemented elements are complemented. Hence $\neg \neg a\vee \neg \neg b$ is regular, and it is the least regular upper bound of $a\vee b$. ThereforeThis proves (ii) implies (iii). The identification of the regular open elements with interiors of closures was established in the preceding part.