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

 $\displaystyle S \cup T = \varnothing$ $\iff$ $\displaystyle \neg \exists x: x \in \left ({S \cup T}\right)$ Definition of Empty Set $\displaystyle$ $\iff$ $\displaystyle \forall x: \neg \left ({x \in \left ({S \cup T}\right)}\right)$ De Morgan's Laws (Predicate Logic) $\displaystyle$ $\iff$ $\displaystyle \forall x: \neg \left ({x \in S \lor x \in T}\right)$ Definition of Set Union $\displaystyle$ $\iff$ $\displaystyle \forall x: x \notin S \land x \notin T$ De Morgan's Laws: Conjunction of Negations $\displaystyle$ $\iff$ $\displaystyle S = \varnothing \land T = \varnothing$ Definition of Empty Set

$\blacksquare$

Proof 2

Let $S \cup T = \varnothing$.

We have:

 $\displaystyle S$ $\subseteq$ $\displaystyle S \cup T$ Set is Subset of Union $\displaystyle \implies \ \$ $\displaystyle S$ $\subseteq$ $\displaystyle \varnothing$ by hypothesis
$\varnothing \subseteq S$

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

Similarly for $T$.

$\blacksquare$