Free ultrafilters on $\mathbb N$ I need to show that if $U$ is a free ultrafilter on $\mathbb N$, then ($A \in U$ and $F \subseteq \mathbb N$ finite) $\implies (A\Delta F) \in U$.  
To show this do you need maximality?  (Would the same hold for any non-principal/free filter; one such that the intersection of all its sets is empty?)
 A: The complement $F'$ of our finite set is in $U$, because $U$ is non-principal. It follows that $A\cap F'\in U$. But the symmetric difference contains $A\cap F'$. We have not used maximality, except for the assertion that the complement of any $1$-element set is in the filter.
A: It's a nice result that an ultrafilter on an infinite set $X$ is non-principal if and only if every cofinite subset of $X$ is a member of the ultrafilter. Consequently, for every finite $F\subseteq\Bbb N$, we have $(\Bbb N\smallsetminus F)\in U$, so $A\cap(\Bbb N\smallsetminus F)\in U$, and since $A\cap(\Bbb N\smallsetminus F)\subseteq A\Delta F$, then $(A\Delta F)\in U$.
We don't need that whole result--only that non-principality of a filter implies that every cofinite subset of $\Bbb N$ is a member of the filter. If we can prove that this holds for arbitrary filters (not just ultrafilters), then maximality won't be necessary at all. I suspect, however, that this may not hold in general. I'll think on it and see if I can come up with another way to get there avoiding maximality.
