Interior of Closure is Regular Open

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H \subseteq S$.

Then $H^{- \circ}$ is regular open.

Proof
We must show that $H^{- \circ - \circ} = H^{- \circ}$.

First we show that $H^{- \circ - \circ} \subseteq H^{- \circ}$.

By the definition of interior, we have $H^{- \circ} \subseteq H^-$.

By Closure of Subset, we have $H^{- \circ -} \subseteq H^{--} = H^-$.

Applying Interior of Subset gives us $H^{- \circ - \circ } \subseteq H^{- \circ}$.

Next we show $H^{- \circ - \circ} \supseteq H^{- \circ}$.

Again using the definition of interior, we have $H^{- \circ -} \supseteq H^{- \circ}$.

By Interior of Subset we have $H^{- \circ - \circ} \supseteq H^{- \circ \circ} = H^{- \circ}$.

Thus by definition of set equality we have demonstrated $H^{- \circ - \circ} = H^{- \circ}$.