Nonprincipal ultrafilter

A nonprincipal ultrafilter contains no finite set. On an infinite set it contains the cofinite filter.

If $U$ is a nonprincipal $\kappa$-complete ultrafilter on $X$ and $A\subseteq X$ has cardinality below $\kappa$, then $A\notin U$. Indeed, every $X\setminus \{a\}$ belongs to $U$, so $\kappa$-completeness puts their intersection $X\setminus A$ in $U$.