Union is Empty iff Sets are Empty
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Theorem
If the union of two sets is the empty set, then both are themselves empty:
- $S \cup T = \O \iff S = \O \land T = \O$
Set of Sets
Let $\SS$ be a set of sets.
Then:
- $\ds \bigcup \SS = \O \iff \forall S \in \SS: S = \O$
Proof 1
| \(\ds \) | \(\) | \(\ds S \cup T = \O\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \neg \exists x: \, \) | \(\ds \) | \(\) | \(\ds x \in \paren {S \cup T}\) | Definition of Empty Set | |||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds \neg \paren {x \in \paren {S \cup T} }\) | De Morgan's Laws (Predicate Logic) | |||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds \neg \paren {x \in S \lor x \in T}\) | Definition of Set Union | |||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds x \notin S \land x \notin T\) | De Morgan's Laws: Conjunction of Negations | |||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(\) | \(\ds S = \O \land T = \O\) | Definition of Empty Set |
$\blacksquare$
Proof 2
Let $S \cup T = \O$.
We have:
| \(\ds S\) | \(\subseteq\) | \(\ds S \cup T\) | Set is Subset of Union | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds S\) | \(\subseteq\) | \(\ds \O\) | by hypothesis |
From Empty Set is Subset of All Sets:
- $\O \subseteq S$
So it follows by definition of set equality that $S = \O$.
Similarly for $T$.
$\blacksquare$