Help to understand a proof (about filters) Niels Diepeveen claims that he has proved the following theorem:
Theorem Consider the cofinite filter $F$ on an infinite set. Let $K$ be a collection of ultrafilters such that $\bigcap K=F$. Then for every $G\in K$ we have $\bigcap (K\setminus\{G\})=F$.
The proof is in this USENET post.
I do not understand the proof, see this my USENET message.
I ask to provide me with full detailed proof based on the Niels Diepeveen's proof or any other alternative proof.
 A: The claim "$\bigcap K$ is cofinite if and only if for every infinite $A$ there exists $G\in K$ with $A\in G$" is false. More precisely, forward direction is true, but the backward direction is wrong.
For instance, let $K$ be the set of all principal ultrafilters on $\omega$. It satisfies a stronger version of the given condition: if $A\subset\omega$ is nonempty, then there is $G\in K$ such that $A\in G$. Yet, $\bigcap K$ is the trivial filter; for, if $X\subset \omega$ is proper, then let $y\in \omega\setminus X$ and note that $X$ is not a member of the principal ultrafilter generated by $y$.
Edit As Niels Diepeveen says in a comment to this answer, the following modified statement is valid and as a result the rest of the proof goes through.
Let $K$ be a collection of free ultrafilters. The intersection $\bigcap K$ is cofinite if and only if for every infinite set $A$, there exists $G\in K$ such that $A\in G$.
Proof. Suppose $\bigcap K$ is cofinite, and let $A$ be a set such that no $G$ in $K$ contains $A$. Then, every member of $K$ contains the complement of $A$. Thus, the complement of the complement of $A$--i.e. $A$--is finite. 
For the other direction, suppose $A$ is an infinite set in $\bigcap K$. If $A'$ is the complement of $A$ and is infinite, then note that $A'\in G$ for some $G\in K$. But this means $A\not\in G\supset\bigcap K$, a contradiction. q.e.d.
