Finite Union of Regular Closed Sets is Regular Closed

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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 $T$

$\blacksquare$


Sources