Date: Nov 7, 2013 2:08 PM
Author: Victor Porton
Subject: Re: Partition of a filter
William Elliot wrote:

> 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?

3.6 Partitioning

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