Problem: An open subset $U$ in a topological space is regular open if $U$ is the interior of its closure. A closed subset is regularly closed if it is the closure of its interior. Show that the complement of a regularly open set is regularly closed and vice versa.
Solution: Let $U$ be any set. We need to show that $U$ is regular open if $U^c$ is regular closed, and $U$ is regular closed if $U^c$ is regular open. Let us first see relations between interior, closure and complement.
For any set $U$, we have
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$(\mathrm{int} (U))^c = \overline{U^c} $ and
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$\left( \bar{U} \right)^c = \mathrm{int}\left( U^c \right) $.
We recall the topological definition of $\bar{U} $ and $\mathrm{int}(U) $.
\begin{align*}
\mathrm{int}(U) & \coloneqq \bigcup \left\{ G: G \subseteq U \text{ and $G$ is open} \right\}, \\
\bar{U} & \coloneqq \bigcap \left\{ F: U \subseteq F \text{ and $F$ is closed} \right\} .
\end{align*}
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We have
\begin{align*}
\left( \mathrm{int}(U) \right) ^c & = \left( \bigcup \left\{ G: G \subseteq U \text{ and $G$ is open} \right\} \right) ^c \\
& = \bigcap \left\{ G^c: G^c \supseteq U^c \text{ and $G^c$ is closed} \right\} = \overline{U^c}.
\end{align*}
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Similarly,
\begin{align*}
\left( \bar{U} \right) ^c & = \left( \bigcap \left\{ F: U \subseteq F \text{ and $F$ is closed} \right\} \right) ^c \\
& = \bigcup \left\{ F^c: F^c \subseteq U^c \text{ and $F^c$ is open} \right\} = \mathrm{int} \left( U^c \right)
\end{align*}
Set $U$ be regular open, that is, $\mathrm{int} (\bar{U} ) = U$. Using the above results, we have
\[
U^c = \left( \mathrm{int} (\bar{U} ) \right) ^c \overset{(1)}{=} \overline{\left( \bar{U}\right)^c } \overset{(2)}{=} \overline{\mathrm{int} \left( U^c \right) }.
\]
Similarly, if $U$ is regular closed, that is, $\overline{\mathrm{int}(U) } = U $, then
\[
U^c = \left( \overline{\mathrm{int}(U) } \right) ^c \overset{(2)}{=} \mathrm{int} \left( \mathrm{int}(U)^c \right) \overset{(1)}{=} \mathrm{int} \left( \overline{U^c} \right).
\]