Solution
= Solution
An <ultrafilter> $U$ on $X$ is <kappa-complete> if, for every index set $J$ with $|J|<\kappa$ and every family $(A_j)_{j\in J}$ of members of $U$,
$$
\boxed{\bigcap_{j\in J}A_j\in U.}
$$
The bound is strictly fewer than $\kappa$ sets. In particular every <ultrafilter> is $\aleph_0$-complete, since this requires only finite intersections; <countably complete ultrafilters> require intersections of countably many members, equivalently $\aleph_1$-completeness.