Intersection is Largest Subset/Family of Sets

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

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

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

Proof
Let $X \subseteq S_i$ for all $i \in I$.

Then:

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

From Intersection is Subset: Family of Sets we have:
 * $\forall i \in I: \bigcap_{j \mathop \in I} S_j \subseteq S_i$

So from Subset Relation is Transitive, it follows that:
 * $\displaystyle \forall i \in I: X \subseteq \bigcap_{j \mathop \in I} S_j \subseteq S_i$

So it follows that $\forall i \in I: X \subseteq S_i$.

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

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

Also see

 * Union is Smallest Superset: Family of Sets