Another AC anomaly?
in [Math]
|
Prev: Dumb Question on Inference For Regression (Ho:= No Linear Relation)
Next: Is this a valid statement?
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
From: Virgil on 18 Dec 2009 21:16 In article <b8a6551e-b171-46c9-8e3b-3de00eb350d8(a)v25g2000yqk.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 17 Dez., 22:47, Quaestor <quaes...(a)qqq.com> wrote: > > > > Please prove your assertion by showing which path I did not produce. > > > But prove it by knowing only all nodes of all paths. > > > > Until you give some explicit way of telling which paths you do produce, > > ther is no way of telling which ones you miss. > > I produce every path. No you don't. In particular, you do not produce any path which contains both infinitely many left branchings and infinitely many right branchings, but there are most common kind in any ACTUAL complete infinite binary tree. > There is no node in the complete binary tree > that is not member of an infinitude of paths. That is still true for demonstrably incomplete sets of paths, as well as many sets of finite sub-paths, so is totally irrelevant to the issue. > > > > > Information exceeding the infinite sequence of nodes of each path is > > > not acceptable in mathematics. > > > > I do not accept your fiats on what is or is not acceptable in > > mathematics, > > In mathematics, numbers are defined by digits. WRONG! Numerals can be defined by digits but numbers can be defined in a number of ways which do not require digits at all. For example, the number pi is standardly defined with no reference to digits at all. And so is the number e. > Additional information > is not part of mathematics. If you do not accept that, then there is > no common basis for discussion. There seems to be no basis for anyone with a reasonable knowledge of mathematics for to have a discussion of mathematics with someone like WM who ignores so much mathematics in order to support his delusions about mathematics. > > My claim is: The set of numbers which can unambiguously be identified > by digit sequences is countable. But in asddition, you claim that these are all of the numbers there are, even though such numbers as e and pi have no known sequence of digits which unambiguously identify them, so cannot be a part of WM's odd version of mathematics. > > > > > > Which path do you see in the tree that has not that property? > > > > In my tree, none. But in there are sets of nodes satisfying my condition > > of pathhood which do not occur in your tree. > > Pure assertion. Give an example, please. In my tree, each path represents a real number between 1 and 2 by inserting a binary radix point after the root node and representing each node of form 2*n + i, i in {0,1}, by its i value. Then, for example, pi/2 corresponds to a path in my tree but not in yours.
From: Virgil on 18 Dec 2009 21:22
In article <3678078d-7bde-4457-896a-2d55ee158088(a)y24g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 17 Dez., 22:52, Quaestor <quaes...(a)qqq.com> wrote: > > > > > > 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, ...}. > > > > > > As usual you confuse to be an element and to be a party of... > > > > > I am concerned with numbers. Your absurd distinctions are irrelevant. > > > > They are neither absurd nor irrelevant to mathematics. > > Ok, then you have a different idea of mathematics. In common > mathematics, numbers can be defined by digit sequences. Not all of them. Negatives require a negative sign , which is not a digit. Decimal numbers reqire a decimal point, which is not a digit. In fact, the only numbers that can be defined by digits alone are the non-negative integers > > > > > > Either N exists or not. If yes, then it is the union of all natural > > > numbers as well as the union of all initial segments of the ordered > > > set of all natural numbers as well as the union of all these and > > > itself, because > > > > > {1, 2, 3, ...} U {1, 2, 3, ...} = {1, 2, 3, ...} > > > > > If you see that this equation is true, then you acknowledge my > > > argument. > > > > On the contrary, that equation is true > > Fine, Virgil, that is one big step to comprehension. But however hard WM tries, he has not what is needed to attain it. |