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Topic:
Product, Filters and Quantales
Replies:
31
Last Post:
Oct 21, 2013 7:52 AM




Re: Product, Filters and Quantales
Posted:
Oct 19, 2013 10:59 PM


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)})] ?
/\_(r>0) (D xx F_{(r,oo)}) = D xx /\_(r>o) F_{(r,oo)} . . = D xx {R}
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}
K in (D xx F_{0}) o (D xx F_{(r,oo)}) . . 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.
 On Sat, 19 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 }).
If F and the Gk's are filters for products, then . . F o inf_k Gk = inf{ F o Gk  k in K }
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). 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)})] ?
> >> 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.
> >> >> > 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}
>> (D xx F_{0}) o /\_(r>0) (D xx F_{(r,oo)}) = D xx F_{0}
Here we differ.
> >> >> > 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. > >> > >



