Let $\mathcal{F} \subseteq \mathcal{P}(\mathbb{N})$ a "described", countable family of subset of $\mathbb{N}$ with the strong finite intersection property (that is, every nonempty finite subfamily of $\mathcal{F}$ has nonempty intersection, and that intersection contains infinite elements.)
With "described" family I mean that $\mathcal{F}$ can be defined as $\mathcal{F}=$ { $A_{n} : n \in \mathbb{N}$ } where, for each $n \in \mathbb{N}$, $A_{n} =$ {$m \in \mathbb{N} : p(m, n) $} and $p(m, n)$ is a "formula" with $m$ and $n$ as parameters, like, for example, $m > n$.
Consider now the filter $f$ generated by $\mathcal{F}$. I claim this filter cannot be an ultrafilter.
Indeed, suppose $f$ is an ultrafilter. Since no finite subset of $\mathbb{N}$ is contained in $f$, we can deduce that for every cofinite $C \subseteq \mathbb{N}$ we have $C \in f$. In other words, $f$ contains the Frèchet filter on $\mathbb{N}$ and this in turn means that $f$ is free.
But this means that we have found a way to describe a free ultrafilter, which, as far as I understand, is impossible.
So $f$ can't be an ultrafilter.
Is the reasoning correct?