# 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:

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

## 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 $T$.

$\blacksquare$