# Finite Intersection of Regular Open Sets is Regular Open

## Theorem

Let $T$ be a topological space.

Let $n \in \N$.

Suppose that:

$\forall i \in \left[{1 \,.\,.\, n}\right]: H_i \subseteq T$

where all the $H_i$ are regular open in $T$, i.e.:

$\forall i \in \left[{1 \,.\,.\, n}\right]: H_i = H_i^{- \circ}$

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

That is:

$\displaystyle \bigcap_{i \mathop = 1}^n H_i = \left({\bigcap_{i \mathop = 1}^n H_i}\right)^{- \circ}$

## Proof

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

$\blacksquare$