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

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  • $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Jul 19 '14 at 18:35
  • $\begingroup$ @PatrickDaSilva $U$ is a set. It does not "have poset structure". $\endgroup$ – porton Jul 19 '14 at 18:39
  • $\begingroup$ 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? $\endgroup$ – Patrick Da Silva Jul 20 '14 at 7:00
  • $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Jul 20 '14 at 8:02
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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 $C_n$.

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