On Wed, 6 Nov 2013, Victor Porton wrote: > William Elliot wrote: > > 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: > > http://www.mathematics21.org/algebraic-general-topology.html
Not possible. Where, in the old numbering is the definition?
> (I fact I define it in two different ways, which are not equivalent.)