Interior of Finite Intersection equals Intersection of Interiors

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Theorem

Let $T$ be a topological space.

Let $n \in \N$.

Let:

$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$


Then:

$\ds \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ = \bigcap_{i \mathop = 1}^n H_i^\circ$

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


Proof

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

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

$\blacksquare$


Sources