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Re: Product, Filters and Quantales
Posted:
Oct 17, 2013 1:18 PM


William Elliot wrote:
> On Wed, 16 Oct 2013, Victor Porton 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)})]
So not.
> /\_(r>0) (D xx F_{(r,oo)}) = D xx /\_(r>o) F_{(r,oo)} > . . = D xx {R}
/\_(r>0) (D xx F_{(r,oo)}) = D xx (0;oo) = D xx /\_(r>o) F_{(r,oo)}
What is {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}
What is {R}?
> K in (D xx F_{0}) o (D xx F_{(r,oo)})
(D xx F_{0}) o (D xx F_{(r,oo)}) = 0
Every set 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.



