Finite Intersection of Regular Open Sets is Regular Open
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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 open in $T$.
That is:
- $\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{- \circ}$
where $H_i^{- \circ}$ denotes the interior of the closure of $H_i$.
Then $\ds \bigcap_{i \mathop = 1}^n H_i$ is regular open in $T$.
That is:
- $\ds \bigcap_{i \mathop = 1}^n H_i = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}$
Proof
\(\ds \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}\) | \(=\) | \(\ds \paren {T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^\circ}^\circ\) | Complement of Interior equals Closure of Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^\circ\) | De Morgan's Laws: Difference with Intersection | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {T \setminus \bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^\circ\) | Complement of Interior equals Closure of Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^-}^\circ\) | De Morgan's Laws: Difference with Intersection | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcap_{i \mathop = 1}^n H_i^-}^\circ\) | Relative Complement of Relative Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcap_{i \mathop = 1}^n H_i^{- \circ}\) | Interior of Finite Intersection equals Intersection of Interiors | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcap_{i \mathop = 1}^n H_i\) | as all $H_i$ are regular open in $T$ |
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors