```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
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