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Topic: Invariance of Sup
Replies: 23   Last Post: Feb 16, 2014 1:58 PM

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@less@ndro

Posts: 215
Registered: 12/13/04
Re: Invariance of Sup
Posted: Feb 16, 2014 1:58 PM
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William Elliot <marsh@panix.com> wrote:
> On Thu, 13 Feb 2014, quasi wrote:
> > William Elliot wrote:
> > >
> > >Let X,Y be two lattices and f:X -> Y a surjective lattice map,
> > >i.e., for all x,y,
> > > f(x inf y) = f(x) inf f(y), f(x sup y) = f(x) sup f(y),
> > >or with shorthand,
> > > for all x,y, f(xy) = f(x).f(y), f(x + y) = f(x) + f(y).
> > >
> > >If X is a complete lattice, is Y a complete lattice?

> >
> > Not necessarily. In the article
> > Quotients of Complete Ordered Sets
> > Maurice Pouzet & Ivan Rival
> > Algebra Universalis, Vol 17 (1983), pages 393-405
> > the authors present this example ...
> >
> > Let N denote the set of natural numbers.
> > Let X be the power set of N.

>
> > Define a congruence on X by
> > A ~ B if (A - B) U (B - A) is finite.

>
> > Let Y = X/~. Let f be the natural map from X onto Y.
>
> What map is that?
> f:P(N) -> P(N)/~, A -> { B | A ~ B }?


Yes. Actually, that counterexample is also Exercise V-1-9 in
Birkhoff's Lattice Theory.

The above relation is in fact an example of a general construction:
let X be a sup-semilattice and D be an ideal of X
(i.e. D is downwards closed and closed under binary sup) and define

a ~ b iff there exists a d in D with a \/ d = b \/ d.

Then ~ is an equivalence and also a congruence w.r.t. \/.
If X happens to be a _distributive_ lattice then ~ is also a
congruence w.r.t. /\.

Now the finite subsets form an ideal in P(N) and for
subsets A and B of N you have

(A \ B) U (B \ A) finite
<==> there is a finite subset F with A U F = B U F

--
Marc



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