Dynkin System Contains Empty Set

Theorem
Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.

Then the empty set $\O$ is an element of $\DD$.

Proof
As $\DD$ is a Dynkin system, $X \in \DD$.

By Set Difference with Self is Empty Set, $X \setminus X = \O$.

Hence, by property $(2)$ of a Dynkin system, $\O = X \setminus X \in \DD$.