# Union is Empty iff Sets are Empty

## Theorem

If the union of two sets is the empty set, then both are themselves empty:

$S \cup T = \O \iff S = \O \land T = \O$

## 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
$\O \subseteq S$

So it follows by definition of set equality that $S = \O$.

Similarly for $T$.

$\blacksquare$