# 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 $H$.

 $\displaystyle \left({\bigcap_{H \mathop \in \mathbb H} H}\right)^\circ$ $=$ $\displaystyle T \setminus \left({T \setminus \bigcap_{H \mathop \in \mathbb H} H}\right)^-$ 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:

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

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

$\blacksquare$