Date: Nov 6, 2013 2:36 PM
Author: Victor Porton
Subject: Re: Partition of a filter

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:

(I fact I define it in two different ways, which are not equivalent.)