On Tue, 5 Nov 2013, Victor Porton wrote: > > As for Conjecture 4.153, obviously no, > > a filter cannot be partitioned into ultrafilters because > > all the ultrafilters contain the same top element. > > > > Do you mean this instead? > > > > If F is a filter for S, can F be partitioned > > into ultrafilters for subsets of S? > > I mean that filter can be partitioned into ultrafilters in the REVERSE > order.
What does order have to do with it? A partition of a set S, or a collection of sets like filters are, is a pairwise disjoint collection whose union is S.