Union of Interiors is Subset of Interior of Union

Theorem
Let $T$ be a topological space.

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

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

Then the union of the interiors of the elements of $\H$ is a subset of the interior of the union of $\H$.
 * $\ds \bigcup_{H \mathop \in \H} H^\circ \subseteq \paren {\bigcup_{H \mathop \in \H} H}^\circ $

Proof 1
In the following, $H^-$ denotes the closure of the set $H$.

At this point we note that:
 * $(1): \quad \ds \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \subseteq \bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$

from Closure of Intersection is Subset of Intersection of Closures.

Then we note that:
 * $\ds T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-} \subseteq T \setminus \paren {\paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} $

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

Then we continue:

Proof 2
Let $\mathbb U$ be the collection of all open subsets of $\bigcup \H$.

Then by definition of interior:


 * $\ds \paren {\bigcup_{H \mathop \in \H} H}^\circ = \paren {\bigcup \H}^\circ = \bigcup \mathbb U$

As $\mathbb U$ contains all open subsets of $\bigcup \H$, and $H^\circ$ is open for any $H \in \H$,


 * $\ds \set {H^\circ : H \in \H} \subseteq \mathbb U$

Thus,


 * $\ds \bigcup_{H \mathop \in \H} H^\circ = \bigcup \set {H^\circ : H \in \H} \subseteq \bigcup \mathbb U$

Also see

 * Interior of Union is not necessarily Union of Interiors