Union of Set of Sets when a Set Intersects All

Theorem
Let $F$ be a set of sets.

Let $S$ be a set or class.

Suppose that:
 * $\forall A \in F: A \cap S \ne \varnothing$

Then:
 * $\displaystyle F = \bigcup_{x \mathop \in S} \left\{{A \in F: x \in A}\right\}$

Proof
Suppose that $B \in F$.

Then $B \cap S$ has an element $x_B$.

Thus $B \in \left\{{A \in F: x_B \in A}\right\}$.

By the definition of union:


 * $\displaystyle B \in \bigcup_{x \mathop \in S} \left\{{A \in F: x \in A}\right\}$

Suppose instead that $\displaystyle B \in \bigcup_{x \mathop \in S} \left\{{A \in F: x \in A}\right\}$.

Then by the definition of union, there exists an $x_B \in S$ such that $B \in \left\{{A \in F: x_B \in A}\right\} \subseteq F$.

Thus $B \in F$.

We have shown that $\displaystyle \forall B: \left({B \in F \iff B \in \bigcup_{x \mathop \in S} \left\{{A \in F: x \in A}\right\}}\right)$.

Therefore $\displaystyle F = \bigcup_{x \mathop \in S} \left\{{A \in F: x \in A}\right\}$ by the Axiom of Extension.