Proposition 5

Let $\mathfrak{F}$ be an ultrafilter on a set $X$. If $A$ and $B$ are two subsets of $X$ such that $A \cup B \in \mathfrak{F}$, then either $A \in \mathfrak{F}$ or $B \in \mathfrak{F}$.

If the proposition is false, there exist subsets $A$ and $B$ of $X$ such that $A \notin \mathfrak{F}$ and $B \notin \mathfrak{F}$ and $A \cup B \in \mathfrak{F}$. Let $\mathfrak{G}$ be the set of subsets $M$ of $X$ such that $A \cup M \in \mathfrak{F}$. It is straightforward to check that $\mathfrak{G}$ is a filter on $X$, and $\mathfrak{G}$ is strictly finer than $\mathfrak{F}$, since $B \in \mathfrak{G}$; but this contradicts the hypothesis that $\mathfrak{F}$ is an ultrafilter.