Union is Smallest Superset/Set of Sets

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:

$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$

$\blacksquare$

Also see

• Intersection is Largest Subset