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

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

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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:

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

$\blacksquare$