Intersection of Interiors contains Interior of Intersection

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Theorem

Let $T$ be a topological space.


Let $\mathbb H$ be a set of subsets of $T$.

That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.


Then the interior of the intersection of $\mathbb H$ is a subset of the intersection of the interiors of the elements of $\mathbb H$.

$\displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$


Proof

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

\(\displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ\) \(=\) \(\displaystyle T \setminus \paren {T \setminus \bigcap_{H \mathop \in \mathbb H} H}^-\) Complement of Interior equals Closure of Complement
\(\displaystyle \) \(=\) \(\displaystyle T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-}\) De Morgan's Laws: Difference with Intersection


At this point we note that:

$(1): \quad \displaystyle \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \supseteq \bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$

from Closure of Union contains Union of Closures.


Then we note that:

$\displaystyle T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} \subseteq T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}$

from $(1)$ and Set Complement inverts Subsets.


Then we continue:

\(\displaystyle T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}\) \(=\) \(\displaystyle T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} T \setminus H^\circ}\) Complement of Interior equals Closure of Complement
\(\displaystyle \) \(=\) \(\displaystyle T \setminus \paren {T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} H^\circ} }\) De Morgan's Laws: Difference with Intersection
\(\displaystyle \) \(=\) \(\displaystyle \bigcap_{H \mathop \in \mathbb H} H^\circ\) Relative Complement of Relative Complement

$\blacksquare$


Also see


Sources