Union of Interiors is Subset of Interior of Union
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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 $H$.
| \(\ds \paren {\bigcup_{H \mathop \in \mathbb H} H}^\circ\) | \(=\) | \(\ds T \setminus \paren {T \setminus \bigcup_{H \mathop \in \mathbb H} H}^-\) | Complement of Interior equals Closure of Complement | |||||||||||
| \(\ds \) | \(=\) | \(\ds T \setminus \paren {\paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-}\) | De Morgan's Laws: Difference with Union |
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:
| \(\ds T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}\) | \(=\) | \(\ds T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} T \setminus H^\circ}\) | Complement of Interior equals Closure of Complement | |||||||||||
| \(\ds \) | \(=\) | \(\ds T \setminus \paren {T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} H^\circ} }\) | De Morgan's Laws: Difference with Union | |||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcup_{H \mathop \in \mathbb H} H^\circ\) | Relative Complement of Relative Complement |
$\blacksquare$
Proof 2
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:
- $\ds \bigcup_{H \mathop \in \H} H^\circ = \bigcup \set {H^\circ : H \in \H} \subseteq \bigcup \mathbb U$
$\blacksquare$