Exterior of Finite Union equals Intersection of Exteriors

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Theorem

Let $T$ be a topological space.

Let $n \in \N$.


Let $\forall i \in \closedint 1 n: H_i \subseteq T$.


Then:

$\displaystyle \paren {\bigcup_{i \mathop = 1}^n H_i}^e = \bigcap_{i \mathop = 1}^n H_i^e$

where $H_i^e$ denotes the exterior of $H_i$.


Proof

In the following, $H_i^\circ$ denotes the interior of the set $H_i$.


\(\ds \paren {\bigcup_{i \mathop = 1}^n H_i}^e\) \(=\) \(\ds \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^\circ\) Definition of Exterior
\(\ds \) \(=\) \(\ds \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ\) De Morgan's Laws: Difference with Union
\(\ds \) \(=\) \(\ds \bigcap_{i \mathop = 1}^n \paren {T \setminus H_i}^\circ\) Interior of Finite Intersection equals Intersection of Interiors
\(\ds \) \(=\) \(\ds \bigcap_{i \mathop = 1}^n H_i^e\) Definition of Exterior

$\blacksquare$


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