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 $\mathbb H \subseteq \mathcal P \left({T}\right)$ where $\mathcal P \left({T}\right)$ is the power set of $T$.


Then:

$\displaystyle \left({\bigcap_{H \mathop \in \mathbb H} H}\right)^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$

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


Proof

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

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


At this point we note that:

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

from Closure of Union contains Union of Closures.


Then we note that:

$\displaystyle T \setminus \left({\left({\bigcup_{H \mathop \in \mathbb H} \left({T \setminus H}\right)}\right)^-}\right) \subseteq T \setminus \left({\bigcup_{H \mathop \in \mathbb H} \left({T \setminus H}\right)^-}\right)$

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


Then we continue:

\(\displaystyle T \setminus \left({\bigcup_{H \mathop \in \mathbb H} \left({T \setminus H}\right)^-}\right)\) \(=\) \(\displaystyle T \setminus \left({\bigcup_{H \mathop \in \mathbb H} T \setminus H^\circ}\right)\) Complement of Interior equals Closure of Complement
\(\displaystyle \) \(=\) \(\displaystyle T \setminus \left({T \setminus \left({\bigcap_{H \mathop \in \mathbb H} H^\circ}\right)}\right)\) 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