Union of Set of Sets when a Set Intersects All

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Theorem

Let $F$ be a set of sets.

Let $S$ be a set or class.

Suppose that:

$\forall A \in F: A \cap S \ne \O$


Then:

$\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$


Proof

Suppose that $B \in F$.

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

Thus:

$B \in \set {A \in F: x_B \in A}$

By the definition of union:

$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$


Suppose instead that:

$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$

Then by the definition of union, there exists an $x_B \in S$ such that:

$B \in \set {A \in F: x_B \in A} \subseteq F$

Thus $B \in F$.


We have shown that:

$\ds \forall B: \paren {B \in F \iff B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A} }$

Therefore by definition of set equality:

$\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$

$\blacksquare$