 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 $U$,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.