Set is Subset of Union/Set of Sets
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Theorem
Let $\mathbb S$ be a set of sets.
Then:
- $\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
Proof
Let $T$ be any element of $\mathbb S$.
We wish to show that $T \subseteq S$.
Let $x \in T$.
Then:
\(\ds x\) | \(\in\) | \(\ds T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcup \mathbb S\) | Definition of Set Union |
Since this holds for each $x \in T$:
\(\ds T\) | \(\subseteq\) | \(\ds \bigcup \mathbb S\) | Definition of Subset |
As $T$ was arbitrary, it follows that:
- $\forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
$\blacksquare$