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:

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

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 $T$

$\blacksquare$


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