# cofinite filter is intersection of all non-principal ultrafilter

It is clear that Cofinite filter on any infinite set is contained in every non-principal ultrafilter so it is contained in the intersection of all non-principal ultrafilters, but my question is that Is Cofinite filter the intersection of all non-principal ultrafilters?

• In other words, you want to know if every infinite set belongs to some non-principal ultrafilter? – bof Oct 12 '16 at 11:31
• Did you roll a die to decide the tags for this question? – Asaf Karagila Oct 12 '16 at 11:44
• What does this have to do with geometric group theory? (your tags) – Paul Plummer Oct 12 '16 at 13:43
• it is used in the theory of asymptotic cones, which was the key ingredient in proof of Gromov's theorem on groups of polynomial growth – Sunny Rathore Oct 12 '16 at 13:46
• – Martin Sleziak Oct 20 '16 at 14:05

Let $E$ be an infinite set. Suppose that a set $A\subseteq E$ belongs to every non-principal ultrafilter on $E;$ we want to show that $A$ is cofinite.
Assume for a contradiction that $A$ is not cofinite, i.e., the set $B=E\setminus A$ is infinite. Then the infinite set $B$ belongs to some non-principal ultrafilter $\mathcal U$ on $E.$ But then $\mathcal U$ is a non-principal ultrafilter on $E$ which does not contain $A,$ contradicting our assumption that $A$ belongs to every non-principal ultrafilter.
Let $\mathcal F=\{X\subseteq E:\ E\setminus X\text{ is finite}\},$ the cofinite filter on$E.$ Then the collection $\{B\}\cup\mathcal F$ has the finite intersection property, whence $\{B\}\cup\mathcal F\subset\mathcal U$ for some ultrafilter $\mathcal U.$ Since $\mathcal U$ contains $\mathcal F,$ it is non-principal.