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 $T$.

$\blacksquare$