Cantor Confusion
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From: David Marcus on 20 Jan 2007 13:35 Andy Smith wrote: > David Marcus writes > > Andy Smith wrote: > >> So, for example we can think of some > >> representative points located in x > > Perhaps I should have said in {x} 0<=x<=1 then. In {x| 0 <= x <= 1}. > My mental problem with the limit of x^n was that viewed as mapping of > points I couldn't see how you could split [0,1] into [0,1) and 1 by any > sort of limit process - I had a mental image of x^n as a y-axis and x as > the x-axis. At n = 1 it is a line, then progressively curving close to > the x-axis as n increases, but always continuous. Viewed from the > y-axis, there always seemed to be an inverse mapping. You are picturing uniform convergence. In fact, Spivak has that identical picture as Figure 6 in Chapter 24. In some sense, uniform convergence is more natural when dealing with functions. -- David Marcus
From: David Marcus on 20 Jan 2007 13:41 David Bernier wrote: > David Marcus wrote: > > Andy Smith wrote: > >> If x is irrational cos(m! pi x) does not have a limit as m->oo, it is > >> undefined? > > > > Correct. It does not have a limit. So, if x is irrational, then > > > > lim_{m->oo) cos(m! pi x) > > > > is undefined. > > For x = 2exp(1) = 2e, cos(m! pi*(2e)) is close to 1 for several > m>1000. 2e = 2/1 + 2/1 + 2/2! + 2/3! + 2/4! + ... 2/(k!) + ... > > Say m is large. Then (m!)*2/((m-1)!) = 2m is even, and for the > k<= m, (m!)*2/k! is even. But for k = m+1, (m!)*2/(k!) = 2/(m+1), > which gets closer to 0 as m increases. I'm not convinced > that lim_{m->oo} cos(m! pi * (2e)) is undefined. I confess I didn't have a proof in mind when I wrote that the limit is undefined. I withdraw my statement. -- David Marcus
From: mueckenh on 20 Jan 2007 17:29 Andy Smith schrieb: > mueckenh(a)rz.fh-augsburg.de writes > > > >Andy Smith schrieb: > > > >> I > >> > > > > The union of all finite binary trees contains all levels > >> > > > >which can be > >> > > > > enumerated by natural numbers: > >> > > > > > >> > > > > 0 0. > >> > > > > / \ > >> > > > > 1 0 1 > >> > > > > / \ / \ > >> > > > > 2 0 1 0 1 > >> > > > > ............................... > >> > > > > > >> > >> Out of interest, aren't the set of all numbers defined by the union of > >> all paths through a finite binary tree with N levels just all the > >> numbers addressed by the first N bits? If so, why do you bother with > >> the tree construction - does it have some special significance? > > > >The real numbers are represented as infinite paths in the "complete" > >infinite tree. Some even twice. > > > >The union of all finite trees is an infinite tree. > >Every finite tree contains only a finite set of paths. > >The countable union of all paths of the finite trees is therefore the > >countable union of all finite paths. > >The countable union of all finite paths is in the union of all finite > >trees. > >The "complete" tree containing all paths is identical to the union of > >al finite trees, with respect to nodes and edges. > >Identical trees cannot contain different sets of paths. > >Therefore, both trees contain the same set of paths. > >Therefore the "complete" set of all path is countable. > >Therefore the set of all real numbers is countable. > >Therefore ZFC is inconsistent. > > I would have said that the set of all paths in a finite tree of depth N > correspond 1:1 with the address range of N bits. You use N as a natural number? It is usual here to denote the set of all natural numbers by N, a natual number by n. > > An infinite tree corresponds to a number encoded in a countably infinite > set of bits. The nodes of the tree, yes. The edges of the tree too. The paths of the tree, no. The tree illustrates the following problem: 1) For every natural number n there is an natural number m such that m = n+1. In short, with n,m in N: An Em: m = n+1 2) The set N of all n exists actually, i.e., it is impossible to change N by adding any further natural number n. This is equivalent with the statement about finite initial segments. 1) An Em: |{1,2,3,...,m}| = |{1,2,3,...,n}| +1. 2) The set N of all {1,2,3,...,n} exists actually, i.e., it is impossible to change N by adding any further finite initial segment {1,2,3,...,n}. > > Cantor's diagonalisation argument then applies. It does not apply to the paths of the tree with all nodes, because this tree contains the representations of all real numbers of the interval [0, 1]. > But, I think that there > are other reasons for thinking that the reals are uncountable anyway. Which reasons have you in mind? > > But I am not qualified to comment anyway. You are qualified, according to Cantor. Understand some German? Cantor said: "Zum Verständnis der Lehre vom Transfiniten bedarf es keiner gelehrten Vorbereitung in der neueren Mathematik; sie kann für diesen Zweck eher schädlich als nützlich sein." Today, that is more true than ever. Regards, WM
From: mueckenh on 20 Jan 2007 17:33 Franziska Neugebauer schrieb: > Whether a _meaning_ of > > T(1) U T(2) U T(3) U ... (inf-u) > > exists does _not_ depend on such question. It depends on a definition > you have to provide. So please answer the question, what (inf-u) means > and prove that it exists. > > > This is connected > > Whether that is "connected" is irrelevant to a possible definition of > (inf-u). That is not a true statement. Nevertheless: Definition of the infinite union of trees by induction: If the finite tree with n levels exists, the finite tree with n+1 levels exist. The tree with 1 level exists. Proof of existence of the union tree by proofs of A, B, and C: A) Proof of the existence of one infinite path by induction over the indexes {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n}... U ... is the same as the union of the ends of the initial segments {1} U { 2} U {3} U... U {n}... U .. which is N which exists. B) Proof of the existence of all infinite paths: See proof of the existence of all real numbers in [0, 1]. C) Proof of the existence of all paths in the tree being infinite:: All paths of a tree have same length by definition. Regards, WM
From: David Marcus on 20 Jan 2007 18:21
mueckenh(a)rz.fh-augsburg.de wrote: > The tree illustrates the following problem: > 1) For every natural number n there is an natural number m such that m > = n+1. > In short, with n,m in N: An Em: m = n+1 > 2) The set N of all n exists actually, i.e., it is impossible to change > N by adding any further natural number n. > > This is equivalent with the statement about finite initial segments. > 1) An Em: |{1,2,3,...,m}| = |{1,2,3,...,n}| +1. > 2) The set N of all {1,2,3,...,n} exists actually, i.e., it is > impossible to change N by adding any further finite initial segment > {1,2,3,...,n}. So, what's the problem (besides your loose wording)? -- David Marcus |