
Re: Partition of a filter
Posted:
Nov 7, 2013 2:08 PM


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/algebraicgeneraltopology.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.)

