Set Union Preserves Subsets/General Result
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Theorem
Let $\mathbb S, \mathbb T$ be sets of sets.
Suppose that for each $S \in \mathbb S$ there exists a $T \in \mathbb T$ such that $S \subseteq T$.
Then $\bigcup \mathbb S \subseteq \bigcup \mathbb T$.
Proof
Let $x \in \bigcup \mathbb S$.
By the definition of union, there exists an $S \in \mathbb S$ such that $x \in S$.
By the premise, there exists a $T \in \mathbb T$ such that $S \subseteq T$.
By the definition of Subset:
- $x \in T$
Thus by the definition of union:
- $x \in \bigcup \mathbb T$
$\blacksquare$