Union of Set of Sets when a Set Intersects All
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Theorem
Let $F$ be a set of sets.
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:
- $\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
$\blacksquare$