Free filter

A free filter on a set $I$ has empty intersection of all its members. This is equivalent to containing every cofinite subset of $I$: for each $i$ some filter member omits $i$, so upward closure gives $I\setminus \{i\}$, and finite intersections give every cofinite set. A proper free filter on an infinite set contains no finite set.

For a free filter $F$, a subset $S$ belongs to $F$ if and only if every ultrafilter extending $F$ contains $S$. If $S\notin F$, every $A\in F$ meets $I\setminus S$ in an infinite set; otherwise a cofinite restriction of $A$ would be contained in $S$. Adjoining $I\setminus S$ therefore generates a proper free filter. The ultrafilter lemma extends it to a nonprincipal ultrafilter witnessing failure of membership.