# Union is Empty iff Sets are Empty/Proof 2

Jump to navigation
Jump to search

## Theorem

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

## Proof

Let $S \cup T = \varnothing$.

We have:

\(\ds S\) | \(\subseteq\) | \(\ds S \cup T\) | Set is Subset of Union | |||||||||||

\(\ds \implies \ \ \) | \(\ds S\) | \(\subseteq\) | \(\ds \varnothing\) | 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 $T$.

$\blacksquare$