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

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Victor Porton

Posts: 621
Registered: 8/1/05
Re: Order, Filters and Reloids
Posted: Aug 26, 2013 8:14 AM
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William Elliot wrote:

> 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.


The improper filter (which I also call a filter) is P(S) that is it contains
empty 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).


OK.

> 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.


principle -> principal

> 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 }


Correct.

> Propositions. Are the following true? Are there proofs?
> The product of filters is associative.


No, it is associative up to an isomorphism.

I have no formal proof for this, yet.

> The composition of reloids is associative.

Correct. Theorem 7.13 in my book:
http://www.mathematics21.org/algebraic-general-topology.html

> 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.


Yes, theorem 7.22 in my book.

> The composition of two complete reloids is complete.

It is a conjecture.

--
Victor Porton - http://portonvictor.org



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