K_h
Posts:
419
Registered:
4/12/07


Re: Another AC anomaly?
Posted:
Dec 18, 2009 9:07 PM


"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

