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:

http://www.mathematics21.org/algebraic-general-topology.html

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