
Re: Principal Reliods
Posted:
Nov 5, 2013 8:18 AM


William Elliot wrote:
> On Mon, 4 Nov 2013, Victor Porton wrote: >> William Elliot wrote: >> >> > F_A = pricnipal filter for S generated by {A} (A subset S). >> > F(C) = the filter for S generated by C subset P(S). >> > >> > Theorem. If for all j in J, Aj subset S, then /\_j F_Aj = F_(\/_j Aj), >> > The interseciton of principal filters is a principal filter. >> > >> > Do you already have a proof for that theorem? >> > It's a one, or at most, two line proof. >> >> In my book: >> >> Corollary 4.86. \uparrow is an order embedding from Z to P. >> >> (Here Z is a set and P is the corresponding set of principal filter.) > > There is no such in my copy. > > 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.

