Union is Empty iff Sets are Empty/Set of Sets
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Theorem
Let $\SS$ be a set of sets.
Then:
- $\ds \bigcup \SS = \O \iff \forall S \in \SS: S = \O$
Proof
\(\ds \) | \(\) | \(\ds \bigcup \SS = \O\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \neg \exists x: \, \) | \(\ds \) | \(\) | \(\ds x \in \paren {\bigcup \SS}\) | Definition of Empty Set | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds \neg \paren {x \in \bigcup \SS }\) | De Morgan's Laws (Predicate Logic) | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds \neg \paren {\exists S \in \SS : x \in S}\) | Definition of Set Union | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall x, \forall S \in \SS: \, \) | \(\ds \) | \(\) | \(\ds x \notin S\) | De Morgan's Laws: Assertion of Universality | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall S \in \SS: \, \) | \(\ds \) | \(\) | \(\ds S = \O\) | Definition of Empty Set |
$\blacksquare$