Intersection of Interiors contains Interior of Intersection

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 $H$.

At this point we note that:
 * $(1): \quad \displaystyle \left({\bigcup_{H \mathop \in \mathbb H} \left({T \setminus H}\right)}\right)^- \subseteq \bigcup_{H \mathop \in \mathbb H} \left({T \setminus H}\right)^-$

from Closure of Union contains Union of Closures.

Then we note that:
 * $\displaystyle T \setminus \left({\left({\bigcup_{H \mathop \in \mathbb H} \left({T \setminus H}\right)}\right)^-}\right) \supseteq T \setminus \left({\bigcup_{H \mathop \in \mathbb H} \left({T \setminus H}\right)^-}\right)$

from $(1)$ and Complements invert Subsets.

Then we continue: