Union is Empty iff Sets are Empty

Theorem

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

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

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 $T$.

$\blacksquare$