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:

$H^{- \circ} \subseteq H^-$

By Topological Closure of Subset is Subset of Topological Closure:

$H^{- \circ -} \subseteq H^{--} = H^-$

By Interior of Subset:

$H^{- \circ - \circ } \subseteq H^{- \circ}$

$\Box$

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

Again by the definition of interior:

$H^{- \circ -} \supseteq H^{- \circ}$

By Interior of Subset:

$H^{- \circ - \circ} \supseteq H^{- \circ \circ} = H^{- \circ}$

$\Box$

Thus by definition of set equality:

$H^{- \circ - \circ} = H^{- \circ}$

$\blacksquare$

Sources

• 1964: Steven A. Gaal: Point Set Topology: $\text{I}: \ \S 2$: Interior, Exterior, Boundary and Closure