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Re: Product, Filters and Quantales
Posted:
Oct 19, 2013 10:24 AM


William Elliot wrote:
> On Fri, 18 Oct 2013, Victor Porton wrote: >> William Elliot wrote: >> >> >> >> > If C subset P(S), then F(A) is the filter for S on P(S) generated >> >> >> > by C. If A subset S, then F_A = F{{A}) the principal filter >> >> >> > generated by A If F,G are filters, then F xx G = F({ AxB  A in >> >> >> > F, B in G }). >> >> > >> >> >> > To recap from your errors and hard to use notation, is this the >> >> >> > counter example for >> >> >> > . . F o inf_k Gk = inf{ F o Gk  k in K } >> >> >> > where F and the Gk's are filters for products? >> >> >> > >> >> >> > D = F({ (r,r) subset R  0 < r }, the neighborhood filter for 0 >> >> >> > in R. F = D xx F_{0} is a filter for RxR on P(RxR). >> >> >> > >> >> >> > Does G_r = D xx F_{(r,oo)}? >> >> >> > >> >> >> > Is this your counter example? >> >> >> > . . F o /\{ G_r  0 < r } /= /\{ F o G_r  0 < r } >> >> >> Yes. >> >> > >> >> > Does >> >> > (D xx F_{0}) o /\_(r>0) (D xx F_{(r,oo)}) >> >> > . . = /\_(r>0) [(D xx F_{0}) o (D xx F_{(r,oo)})] ? >> >> >> >> (D xx F_{0}) o /\_(r>0) (D xx F_{(r,oo)}) = >> >> (D xx F_{0}) o (D xx F_{(0,oo)}) = >> >> D xx F_{0} != >> >> 0 = >> >> /\_(r>0) [(D xx F_{0}) o (D xx F_{(r,oo)})] >> >> Which of the above equalities is unclear? >> >> >> > /\_(r>0) (D xx F_{(r,oo)}) = D xx /\_(r>o) F_{(r,oo)} >> >> > . . = D xx {R} >> >> Yes. >> >> >> /\_(r>0) (D xx F_{(r,oo)}) = >> >> D xx (0;oo) = >> >> D xx /\_(r>o) F_{(r,oo)} >> >> Yes. >> >> >> What is {R}? >> >> >> > { a,b } = { x  x = a or x = b } >> > {a} = { x  x = a } >> > >> >> > K in (D xx F_{0}) o (D xx {R} >> >> > . . iff some A in DxxF_{0}, B in Dxx{R} } with AoB subset K >> >> > . . iff some U in D, V in F_{0}, W in D with UxV o DxR subset K >> >> > . . iff some U in D with UxR subset K iff K in D xx {R} >> >> > >> >> > (D xx F_{0}) o /\_(r>0) (D xx F_{(r,oo)}) = D xx {R} >> >> No, >> >> (D xx F_{0}) o /\_(r>0) (D xx F_{(r,oo)}) = D o F_{0} > > No, the composition of two reloids is a reloid. > It can't be the composition of two filters for R on P(R) > because composition isn't defined for filters except for > filters for products which D and F_{0} certainly aren't.
Sorry, typo. It should be:
(D xx F_{0}) o /\_(r>0) (D xx F_{(r,oo)}) = D xx F_{0}
>> >> > K in (D xx F_{0}) o (D xx F_{(r,oo)}) >> >> >> >> (D xx F_{0}) o (D xx F_{(r,oo)}) = 0 >> >> >> > What's 0? Is it different than the 0 in R. >> > If so, then don't use it; use some different notation. >> >> I meant 0 = PR >> > "PR"? The acronym for "Public Relations"? > Perhaps you mean P(R), the largest filter for R.
I mean P(R), the largest filter for R.
>> >> > . . iff some A in D xx F_{0}, B in D xx F_{(r,oo)} with AoB subset K >> >> > . . iff some U in D, V in F_{0}, W in D, X in F_{(r,oo)} >> >> > . . . . with UxV o WxX subset K >> >> > . . iff some U in D, X in F_{(r,oo)} with UxX subset K >> >> > . . iff K in D xx F_{(r,oo)} >> >> > >> >> > /\_(r>0) [(D xx F_{0}) o (D xx F_{(r,oo)})] >> >> > . . = /\_(r>0) (D xx F_{(r,oo)}) = D xx /\_(r>0) F_{(r,oo)} = D xx >> >> > {R} >> >> > >> >> > Yes, they're equal. >>



