Filters on a set of filters, are they equivalent to just filters? Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.
Let $U$ be a set. Is $F(F(\mathscr{P}U))$ order isomorphic to $F(\mathscr{P}U)$?
 A: No, they are not isomorphic if $U$ is infinite.  Note first that $F(\mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).
On the other hand, I claim $F(F(\mathscr{P}U))$ does not have this property.  First, the atoms of $F(F(\mathscr{P}U))$ are the maximal filters on $F(\mathscr{P}U)$.  Any maximal filter $M$ on $F(\mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$.  Given an ultrafiler $\omega$ on $U$, let us write $M_\omega$ for the corresponding maximal filter on $F(\mathscr{P}U)$, i.e. atom in $F(F(\mathscr{P}U))$.  Note notice that the join in $F(F(\mathscr{P}U))$ of a collection of atoms $M_{\omega_i}$ is just the principal filter on $F(\mathscr{P}U)$ generated by $\bigcap_i \omega_i$.  So, any non-principal filter on $F(\mathscr{P}U)$ is not a join of atoms.
Examples of non-principal filters on $F(\mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(\mathscr{P}U)$ with the lattice of closed subsets of $\beta U$ (the space of ultrafilters on $U$), ordered by inclusion.  In particular, for instance, if $\omega\in\beta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $\omega$.  Or, you could take any strictly decreasing sequence of closed sets $C_1\supset C_2\supset\dots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.
