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.)