Interior of Finite Intersection equals Intersection of Interiors

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:

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

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

Proof

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

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

$\blacksquare$