> 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.
No, for filters I define it differently. See my book: