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

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William Elliot

Posts: 2,637
Registered: 1/8/12
Order, Filters and Reloids
Posted: Aug 25, 2013 9:52 PM
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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 semi-lattice.

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.

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