# 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

\(\displaystyle S \cup T = \varnothing\) | \(\iff\) | \(\displaystyle \neg \exists x: x \in \left ({S \cup T}\right)\) | 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 |

From Empty Set is Subset of All Sets:

- $\varnothing \subseteq S$

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

Similarly for $T$.

$\blacksquare$