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 $F(\mathscr{P}U)$?

• Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know. – Patrick Da Silva Jul 19 '14 at 18:35
• @PatrickDaSilva $U$ is a set. It does not "have poset structure". – porton Jul 19 '14 at 18:39
• Yes, sorry. I know what you mean, I was just thinking of a more general case (where $\mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise? – Patrick Da Silva Jul 20 '14 at 7:00
• Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $\mathscr P(U) \simeq F(\mathscr P(U))$ so the question becomes boring. Interesting question. – Patrick Da Silva Jul 20 '14 at 8:02

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 $$C_n$$.