# Finite Union of Regular Closed Sets is Regular Closed

## 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 closed in $T$, i.e.:

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

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

That is:

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

## Proof

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

$\blacksquare$