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 $\mathcal{F}$, $\bigcap_{i \in N}{F_{i}} \in \mathcal{F}$?


1 Answer 1


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 $V_\kappa \models \not\exists\text{ measurable cardinal}$.

  • $\begingroup$ Thanks for the answer! Believe it or not, this question appeared in a Statistical problem a friend of mine is working with. From your addition, it is very clear that statistical hypothesis testing is never done on subsets of $\Omega$ with a measurable cardinal. $\endgroup$
    – madprob
    Nov 1, 2012 at 8:48

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