Proposition 7

Every filter $\mathfrak{F}$ on a set $X$ is the intersection of the ultrafilters finer than $\mathfrak{F}$.

Clearly this intersection contains $\mathfrak{F}$. Conversely, let $A$ be a subset of $X$ which does not belong to $\mathfrak{F}$, and let $A'$ denote $\complement A$; $A$ contains no set of $\mathfrak{F}$; hence every $M \in \mathfrak{F}$ meets $A'$ and therefore (no. 2, Proposition 1, Corollary 1) there is a filter $\mathfrak{F}'$ which is finer than $\mathfrak{F}$ and contains $A'$. If $\mathfrak{U}$ is an ultrafilter finer than $\mathfrak{F}'$ (Theorem 1) it follows that $A \notin \mathfrak{U}$. This completes the proof.