Another AC anomaly?
in [Logic]
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From: Dik T. Winter on 18 Dec 2009 09:22 In article <dd852490-e622-4bc4-b858-dfcc3b142f8c(a)l13g2000yqb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: .... > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > No. Suppose we have the paths 0.000 and 0.100, what is their union? > > > > And is it a path? > > > > > No. But the union contains two paths. And an infinite union of that > > > kind may contain the path 0.111... > > > > So according to WM a union of sets may contain an object not contained > > in any of the sets being unioned. > > For paths and initial segments to contain and to be is the same. Eh? paths contain nodes and initial segments contain numbers. But be aware that you use the word 'contains' with two different meanings at different times. > For paths and initial segments to contain and to be is the same. In > fact: > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > and also > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. In the mathematical sense the union contains numbers, not sets. > This is so according to set theory. Of course it is rubbish. Well, if you want to use terminology in a different meaning than standard, of course. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 18 Dec 2009 09:26 In article <jrydnSyLVZ_6vbbWnZ2dnUVZ_vOdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:Kusoo8.xz(a)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. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: K_h on 18 Dec 2009 19:56 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:KuuL76.5GA(a)cwi.nl... > In article <eaCdnfgmKeFaorbWnZ2dnUVZ_hmdnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > ... > > > Under the standard > > > construction each natural is `accumulated' in the > > > limit > > > set, since each set persists beyond its introduction, > > > and > > > this preserves the notion of n growing bigger as one > > > proceeds: |n| is 0,1,2,3,... and is N in the limiting > > > case. > > > > Yo, it should be |N| in the limiting case! > > Eh? With your definition, lim {n} = {N} (as you stated > earlier) and so > lim n = N. But now you actually state (as |n| = n), lim n > = |N|. Are > N and |N| equal or not? Yes, in this case they are equal. > > > proceeds: |n| is 0,1,2,3,... and is N in the limiting > > > case. "N in the limiting case" was a typo, I meant "|N| in the limiting case" but it is correct in either case. This post had nothing to do with my definition. Using just wikipedia's definitions for a limit set and the standard construction of the naturals: lim (n-->oo) n = N //for sets. |lim (n-->oo) n| = lim (n-->oo) |n| = |N| //for cardinals Using the standard construction of the naturals, it is a theorem in ZF that if n is a finite ordinal then |n|=n. There is no contradiction here because aleph_0 is defined to be the first ordinal w=N so the limit ordinal w equals lim(n-->oo) n = N and since |N|=aleph_0=w=N, |N|=N in this case. k
From: K_h on 18 Dec 2009 20:10 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:KuuL11.56J(a)cwi.nl... > In article <SqSdnSUF1q8FobbWnZ2dnUVZ_uCdnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > ... > > With the second option > > we > > can restrict ourselves just to the wikipedia definitions > > and > > define the sets n by: > > > > n = 0 = {} > > n = 1 = {{}} > > n = 2 = {{{}}} > > ... > > > > Wikipedia gives liminf(n-->oo) n = 0. > > Only when we use the definition of limit on sequences of > sets, not when we > use the definition of limit on sequences of numbers. You > have to distinguish > the two and clearly state which one you are using. In ZF set theory numbers are defined by sets and in this case there is no difference: 0 = {} 1 = {{}} 2 = {{{}}} ... If the natural numbers are defined by these sets then liminf(n-->oo)n=0. But yes, on the real number line, the "naturals" are still defined by sets but limits on real numbers are not defined by limits on their associated sets. k
From: K_h on 18 Dec 2009 21:07
"Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:Kuuqrs.FJ7(a)cwi.nl... > In article <jrydnSyLVZ_6vbbWnZ2dnUVZ_vOdnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > > news:Kusoo8.xz(a)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 |