[Comment 0] : Occasionally, one might see condition (1) requiring that both $\emptyset$ and $X$ are in $\mathcal{F}$. However, results R2067: Subtracting empty set from set and R2066: Difference of set with itself show that $X \setminus \emptyset = X$ and $X \setminus X = \emptyset$, so that condition (2) ensures $\emptyset, X \in \mathcal{F}$ if only one of $\emptyset$ or $X$ is assumed to belong to $\mathcal{F}$.