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Topic:
Order, Filters and Reloids
Replies:
11
Last Post:
Sep 1, 2013 10:54 AM




Order, Filters and Reloids
Posted:
Aug 25, 2013 9:52 PM


F is a filter for S when F subset P(S)\{empty set}, F not empty and F ordered by subset is a down directed, upper set.
Since a not empty intersection of filters is again a filter, the set of filters ordered by subset is a complete lower semilattice.
If the empty set is allowed to be in filters, then there's a maximum filter P(S), and the (partially) ordered set of filters is a complete order. In what follows, "filter" will mean a filter in the usual sense or the improper maximum filter P(S).
Let F be a filter for X, G a filter for Y. Then B = { UxV  U in F, G in Y } subset P(XxY) is a filter base and the product of F and G, FxxG, is the filter generated by B.
(F,X,Y) is a reloid when F is a filter for XxY.
(F,X,Y) is a complete reloid when there's some A for which A subset { GxxH  G principle ultrafilter for X, H ultrafilter for Y } and F = /\A, the great intersecion of A, the infinum of A.
The composition of two reloids (F,X,Y) and (G,Y,Z) is the reloid (F,X,Y) o (G,Y,Z) = (H,X,Z) where H is the filter generated by { AoB  A in F, B in G } and the compositon of A and B, AoB = { (x,z) in XxZ  some y in Y with (x,y) in A, (y,z) in B }
Propositions. Are the following true? Are there proofs? The product of filters is associative. The composition of reloids is associative.
If A is a set of filters for X, B a set of filters for Y, then /\{ FxxG  F in A, G in B } = /\A xx /\B.
The composition of two complete reloids is complete.



