Sigma-Algebra Contains Empty Set

Theorem
Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.

Then $\varnothing \in \Sigma$.

Proof
Axiom $(1)$ of a $\sigma$-algebra grants $X \in \Sigma$.

By axiom $(2)$ and Set Difference Self Null, it follows that $\varnothing = X \setminus X \in \Sigma$.