Date: Nov 6, 2013 10:38 PM
Author: William Elliot
Subject: Re: Partition of a filter
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:
Not possible. Where, in the old numbering is the definition?
> (I fact I define it in two different ways, which are not equivalent.)