# Finite Union of Regular Closed Sets is Regular Closed

## Theorem

Let $T$ be a topological space.

Let $n \in \N$.

Suppose that:

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

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

That is:

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

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

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

That is:

$\displaystyle \bigcup_{i \mathop = 1}^n H_i = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}$

## Proof

 $\displaystyle \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}$ $=$ $\displaystyle \paren {T \setminus \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^-}^-$ Complement of Interior equals Closure of Complement $\displaystyle$ $=$ $\displaystyle \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^-$ De Morgan's Laws: Difference with Union $\displaystyle$ $=$ $\displaystyle \paren {\paren {T \setminus \bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^-$ Complement of Interior equals Closure of Complement $\displaystyle$ $=$ $\displaystyle \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^\circ}^-$ De Morgan's Laws: Difference with Intersection $\displaystyle$ $=$ $\displaystyle \paren {\bigcup_{i \mathop = 1}^n H_i^\circ}^-$ 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 $T$

$\blacksquare$