TITLE: Existence of non-trivial ultrafilter closed under countable intersection
QUESTION [8 upvotes]: Under what conditions on $\Omega$ does there exist $\mathcal{F} \subset \mathcal{P}(\Omega)$ such that $\mathcal{F}$ is a non-trivial ultrafilter and, for every sequence $(F_{i})_{i \in N}$ of elements of $\mathcal{F}$, $\bigcap_{i \in N}{F_{i}} \in \mathcal{F}$?

REPLY [9 votes]: This is actually a very strong property.  It is an old result that if $\kappa$ is the least cardinal such that there is a countably complete nonprincipal ultrafilter on $\kappa$ (i.e., the ultrafilter is closed under countable intersections), then that ultrafilter is actually $\kappa$-complete (closed under $\lambda$ intersections for all $\lambda < \kappa$), and therefore $\kappa$ is a measurable cardinal.
Therefore the cardinality of $\Omega$ must be at least the smallest measurable cardinal.
Addition: The main point of this being that measurable cardinals may not exist.  For example, it is a result of Dana Scott that measurable cardinals provably do not exist in Gödel's Constructible Universe.  At a more basic level, as measurable cardinals are strongly inaccessible if $\kappa$ is the least measurable cardinal, then $V_\kappa$ is a model of ZFC and $V_\kappa \models \not\exists\text{ measurable cardinal}$.