# Set Union Preserves Subsets/General Result

## 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:

$x \in \bigcup \mathbb T$

$\blacksquare$