Union is Smallest Superset/Set of Sets

Theorem
Let $T$ be a set.

Let $\mathbb S$ be a set of sets.

Then:


 * $\displaystyle \left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$

Proof
By Union of Subsets is Subset: Set of Sets:


 * $\displaystyle \left({\forall X \in \mathbb S: X \subseteq T}\right) \implies \bigcup \mathbb S \subseteq T$

For the converse implication, suppose that $\displaystyle \bigcup \mathbb S \subseteq T$.

Consider any $X \in \mathbb S$ and take any $x \in X$.

From Subset of Union: Set of Sets we have that $X \subseteq \bigcup \mathbb S$.

Thus $\displaystyle x \in \bigcup \mathbb S$.

But $\displaystyle \bigcup \mathbb S \subseteq T$.

So it follows that $X \subseteq T$.

So:


 * $\displaystyle \bigcup \mathbb S \subseteq T \implies \left({\forall X \in \mathbb S: X \subseteq T}\right)$

Hence:


 * $\displaystyle \left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$

Also see

 * Intersection Largest: Set of Sets