Re: lattice theory

From: Philip Reny (preny_at_uchicago.edu)
Date: 09/27/04 

Date: Mon, 27 Sep 2004 13:30:02 +0000 (UTC)

On Sun, 26 Sep 2004 14:00:09 +0000 (UTC), William Elliot wrote:
>From: Philip reny <preny@uchicago.edu>
>Newsgroups: sci.math.research
>Subject: Re: lattice theory
>
> >William Elliot wrote:
> >>From: Philip Reny <preny@uchicago.edu>
>>
> >>X = [0,1]^N; 0 = 0^N
>>
> >>||x|| = sup_i x_i
>>
> >>x <= y when for all i in N, x_i <= y_i
>
> >>>Assume that:
> >>>(1) (X, ||.||) is compact.
>
> >>>(2) Every pair x,y in X possesses a l.u.b., xVy, in X.
> >>lub x,y = prod_i max x_i,y_i = f:N -> [0,1], i -> max x_i,y_i
>>
> >Not necessarily true, and this is the main sticking point. The lub
> >of {x,y} is, by definition, the smallest member of the set of all
> >upper bounds of {x,y} contained in X. There is no reason for this
to
> >be the componentwise max, since X need not containing the
> >componentwise max of the two vectors x,y.
>
>As [0,1] is linear order, max x_i,y_i is defined.
>Let z_i = max x_i,y_i
>Thus x,y <= z. z is upper bound.
>
>Now if x,y <= a, ie if a is an upper bound,
>then for all i, x_i <= a_i and y_i <= a_i
>Thus I conclude for all i, z_i = max x_i,y_i <= a_i.
>Namely z <= a and subsequently z is the smallest upper bound.
>
>Have you any dispute with this proof?

Yes I dispute this proof. The problem is that z need not be contained
in X and hence does not qualify as an upper bound. You must behave as
if the only points that exist are those in X. To take a simple
example, consider the lattice, X, in R^2 consisting of the four
vertices of the unit square and the center point. That is,
X={(0,0),(0,1),(1,0),(1/2,1/2),(1,1)}. The lub of {(1,0),(1/2,1/2)} is
(1,1), NOT (1,1/2), the componentwise max, because (1,1/2) is not a
member of X.

>
> >>>(3) x_{n} --> x and y_{n} --> y imply x_{n} V y_{n} --> xVy.
>
> >Technically, this means that the semilattice is "topological" in
the
> >sense of, say, Birkhoff's Lattice Theory text. That is, the join
> >operator is continuous.
>
> >>That x_n isn't a component of x but a sequence of points of X ?
> >True.
>
> >>>Prove or disprove:
> >>>x_{n} --> 0 implies x_{n} V x_{n+1} V x_{n+2} V ... --> 0.
>>
> >>As x_n -> 0
> >>for all eps, some n with for all k > n, ||x_k|| < eps.
> >True.
>
> >>X is a complete lattice.
> >True, by (1).
>I'm using 'complete' to mean having all sup's and inf's.
>Are you using 'complete' to mean every Cauchy sequence converges?
>

No, I am using complete to mean that all subsets have a supremum (not
necessarily an inf because we are in a semilattice not a lattice). But
I must correct myslef here. Conditions (1) and (3), not (1) alone,
imply completeness. Of course, compactness alone implies Cauchy
completeness.

> >>Again by associativity of sup, don't I have
> >> ||sup C|| = sup { ||x|| : x in C } ?
> >No. Once again this is because lub is not necessarily the
> >componentwise max.
>
>for C subset X,
> sup C = f:N -> [0,1], k -> sup { x_k | x in C }
> = prod_k sup{ x_k | x in A }
>is component wise and is proven like lub x,y was proven component
wise.
>Shall I detail proof or have you found flaw in the above proof?
>

The flaw is, as above, that X need not contain the componentwise sup.

> >>For all eps, some n with for all k > n, ||x_k|| < eps. Thus
> >> || sup { x_(n+k) | k in N } || <= eps
> >>and with an iota of juggling eps, establish the limit you wish to
> >>show?
>
> >No, for the reasons stated above.
>
>Email: privacy.org = agora.rdrop.com

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