Date: Dec 18, 2009 9:07 PM
Author: K_h
Subject: Re: Another AC anomaly?

"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message

news:Kuuqrs.FJ7@cwi.nl...

> In article <jrydnSyLVZ_6vbbWnZ2dnUVZ_vOdnZ2d@giganews.com>

> "K_h" <KHolmes@SX729.com> writes:

> > "Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message

> > news:Kusoo8.xz@cwi.nl...

> ...

> > > Let's have some arbitrary object 'a' and the natural

> > > numbers. Create

> > > the sequence A_n where A_n = {a} and the sequence B_n

> > > where B_n = {n}.

> > > According to your definition:

> > > lim sup A_n = {a}

> > > and

> > > lim sup B_n = {N}.

> > > Now create the sequence C_n: C_2n = A_n, C_2n+1 = B_n.

> > > Again according

> > > to your definition:

> > > lim sup C_n = {a}

> > > which is not equal to union (lim sup A_n, lim sup

> > > B_n).

> >

> > This is a good example, thanks. Your theorem only

> > applies

> > in special cases for the definition I have offered

> > (although

> > my definition satisfies some different but interesting

> > theorems).

>

> Such as? Certainly not:

> limsup | S_n | = |limsup S_n|

> because see for that the sequence C_n above and limsup.

> Stranger,

> with your definition, lim C_n does exist and is equal to

> {a}, but

> lim B_n equals {N}, where B_n is a subsequence of C_n.

> Strange

> that an infinite subsequence can have a limit different

> from the

> limit of the original sequence.

How about:

Let A_n and B_n be two sequences of sets of the form {X_n}.

Let A_s = lim sup A_n and A_i = lim inf A_n, similar for B_s

and B_i. Let C_n be the sequence defined as C_2n = A_n and

C_(2n+1) = B_n.

Theorem:

Since A_s = {a_s} and B_s = {b_s}

lim sup C_n = {a_s \/ b_s}

lim inf C_n = {a_i /\ b_i}

k