Interior of Finite Intersection equals Intersection of Interiors

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Theorem

Let $T$ be a topological space.

Let $n \in \N$.


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


Then:

$\displaystyle \left({\bigcap_{i \mathop = 1}^n H_i}\right)^\circ = \bigcap_{i \mathop = 1}^n H_i^\circ$

where $H_i^\circ$ denotes the interior of $H_i$.


Proof

In the following, $H_i^-$ denotes the closure of the set $H_i$.


\(\displaystyle \left({\bigcap_{i \mathop = 1}^n H_i}\right)^\circ\) \(=\) \(\displaystyle T \setminus \left({T \setminus \bigcap_{i \mathop = 1}^n H_i}\right)^-\) Complement of Interior equals Closure of Complement
\(\displaystyle \) \(=\) \(\displaystyle T \setminus \left({\left({\bigcup_{i \mathop = 1}^n \left({T \setminus H_i}\right)}\right)^-}\right)\) De Morgan's Laws: Difference with Intersection
\(\displaystyle \) \(=\) \(\displaystyle T \setminus \left({\bigcup_{i \mathop = 1}^n \left({T \setminus H_i}\right)^-}\right)\) Closure of Finite Union equals Union of Closures
\(\displaystyle \) \(=\) \(\displaystyle T \setminus \left({\bigcup_{i \mathop = 1}^n T \setminus H_i^\circ}\right)\) Complement of Interior equals Closure of Complement
\(\displaystyle \) \(=\) \(\displaystyle T \setminus \left({T \setminus \left({\bigcap_{i \mathop = 1}^n H_i^\circ}\right)}\right)\) De Morgan's Laws: Difference with Intersection
\(\displaystyle \) \(=\) \(\displaystyle \bigcap_{i \mathop = 1}^n H_i^\circ\) Relative Complement of Relative Complement

$\blacksquare$


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