Union is Smallest Superset/Set of Sets
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Theorem
Let $T$ be a set.
Let $\mathbb S$ be a set of sets.
Then:
- $\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
Proof
By Union of Subsets is Subset: Set of Sets:
- $\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
$\Box$
For the converse implication, suppose that $\ds \bigcup \mathbb S \subseteq T$.
Consider any $X \in \mathbb S$ and take any $x \in X$.
From Set is Subset of Union: Set of Sets we have that $X \subseteq \bigcup \mathbb S$.
Thus $\ds x \in \bigcup \mathbb S$.
But $\ds \bigcup \mathbb S \subseteq T$.
So it follows that $X \subseteq T$.
So:
- $\ds \bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$
$\Box$
Hence:
- $\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
$\blacksquare$