Union of Interiors is Subset of Interior of Union/Proof 2

Proof
Let $\mathbb U$ be the set 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:


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