Question about filters. Conceptually, I am thinking that the existence of an initial finite set on which the entire filter $F$ is built means that every other arbitrary set $B\in F$ contains the initial set $A$. Then it would be true that $n \in C$ for every $C \in F$. But I am not sure that the statement "$A \in F$ and $B \subseteq \mathbb{N}$ with $A \subseteq B$ implies $B \in F$" necessarily gives that for every $B \in F$, $A \subseteq B$.
 A: Here is what you could do for a two-element set: if $\{a,b\}$ is contained in the ultrafilter, then either $\{a\}$ is in the ultrafilter or $\{a\}^c$, the complement, is in the ultrafilter. Now use that filters are closed under intersections to obtain that if $\{a\}^c$ is in the ultrafilter, so is $\{b\}$. Now you can just use induction.
A: Since a filter does not contain the empty set and is closed under intersections, no filter can contain a finite set and all cofinite sets. Since a non-principal ultrafilter is free, the result follows from the following simple lemma:
Lemma: A filter $\mathcal{F}$ is free if and only if it contains the filter of cofinite sets.
Proof: Clearly, containing all cofinite sets is sufficient for a filter to be free. Conversely, let $\mathcal{F}$ be a free filter and let $F$ be a finite set. Since $\mathcal{F}$ is free, there is for each $x$ in $F$ a set $F_x\in\mathcal{F}$ such that $x\notin F_x$. Then $F^C\supseteq\bigcap_{x\in F}F_x\in\mathcal{F}$ and hence $F^C\in\mathcal{F}$.
