# Finite Intersection of Regular Open Sets is Regular Open

## Theorem

Let $T$ be a topological space.

Let $n \in \N$.

Suppose that:

$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$

where all the $H_i$ are regular open in $T$.

That is:

$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{- \circ}$

where $H_i^{- \circ}$ denotes the interior of the closure of $H_i$.

Then $\ds \bigcap_{i \mathop = 1}^n H_i$ is regular open in $T$.

That is:

$\ds \bigcap_{i \mathop = 1}^n H_i = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}$

## Proof

 $\ds \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}$ $=$ $\ds \paren {T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^\circ}^\circ$ Complement of Interior equals Closure of Complement $\ds$ $=$ $\ds \paren {T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^\circ$ De Morgan's Laws: Difference with Intersection $\ds$ $=$ $\ds \paren {\paren {T \setminus \bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^\circ$ Complement of Interior equals Closure of Complement $\ds$ $=$ $\ds \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^-}^\circ$ De Morgan's Laws: Difference with Intersection $\ds$ $=$ $\ds \paren {\bigcap_{i \mathop = 1}^n H_i^-}^\circ$ Relative Complement of Relative Complement $\ds$ $=$ $\ds \bigcap_{i \mathop = 1}^n H_i^{- \circ}$ Interior of Finite Intersection equals Intersection of Interiors $\ds$ $=$ $\ds \bigcap_{i \mathop = 1}^n H_i$ as all $H_i$ are regular open in $T$

$\blacksquare$