Union is Smallest Superset/Family of Sets

Theorem
Let $\left \langle{S_i}\right \rangle_{i \in I}$ be a family of sets indexed by $I$.

Then for all sets $X$:
 * $\displaystyle \left({\forall i \in I: S_i \subseteq X}\right) \iff \bigcup_{i \mathop \in I} S_i \subseteq X$

where $\displaystyle \bigcup_{i \mathop \in I} S_i$ is the union of $\left \langle{S_i}\right \rangle$.

Proof
Suppose that $\forall i \in I: S_i \subseteq X$.

Consider any $\displaystyle x \in \bigcup_{i \mathop \in I} S_i$.

By definition of set union, it follows that:
 * $\exists i \in I: x \in S_i$

But as $S_i \subseteq X$ it follows that $x \in X$.

Thus it follows that:
 * $\displaystyle \bigcup_{i \mathop \in I} S_i \subseteq X$

So:
 * $\displaystyle \left({\forall i \in I: S_i \subseteq X}\right) \implies \bigcup_{i \mathop \in I} S_i \subseteq X$

Now suppose that $\displaystyle \bigcup_{i \mathop \in I} S_i \subseteq X$.

Consider any $i \in I$ and take any $x \in S_i$.

From Subset of Union: Family of Sets we have that:
 * $\displaystyle S_i \subseteq \bigcup_{i \mathop \in I} S_i$

Thus:
 * $\displaystyle x \in \bigcup_{i \mathop \in I} S_i$

But:
 * $\displaystyle \bigcup_{i \mathop \in I} S_i \subseteq X$

So it follows that $S_i \subseteq X$.

So:
 * $\displaystyle \bigcup_{i \mathop \in I} S_i \subseteq X \implies \left({\forall i \in I: S_i \subseteq X}\right)$

Hence:
 * $\displaystyle \left({\forall i \in I: S_i \subseteq X}\right) \iff \bigcup_{i \mathop \in I} S_i \subseteq X$

Also see

 * Intersection Largest/Family of Sets