Cantor Confusion
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From: William Hughes on 16 Jan 2007 09:35 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > In article <1168779834.276761.153130(a)a75g2000cwd.googlegroups.com>, > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > > > > Recall Cantor's diagonal. It consists only of finite initial segments. > > > > > > > > Hardly. > > > > > > Which digit marks an infinite initial segment? > > > > Given an infinite (in your terms potentially infinite) sequence of > > digits > > (e.g Cantor's diagonal), there is one initial segment that is > > not "marked by a digit". This is the sequence (in your terms > > potentially infinite sequence) of all digits. > > Given the sequence of initial segments of binary 1/3: > 0.0 > 0.01 > 0.010 > 0.0101 > ... No, this is not a list of all the initial segments of binary 1/3. You are missing one. The (potentially) infinite sequence of all digits is not in this list. > there is one element not marked by a digit, i.e., not contained in a > finite binary tree but contained in the union of all finite binary > trees. No. The (potentially) infinite sequence of all digits is not in the list, nor is it in the union of all finite binary trees. > > > Thus one cannot say > > Cantor's diagonal "consists only of finite initial segments". > > One must say so. It is similar to saying N consists of finite numbers > only. No. The statement All elements of N are finite numbers is true. The statement All initial segments of Cantor's diagonal are finite is false. There is one (potentially) infinite initial segment. - William Hughes
From: mueckenh on 16 Jan 2007 09:37 Virgil schrieb: > The union of all such restricted infinite binary trees will be a subtree > of the compete infinite binary tree. > > So far, I hope WM and I agree. Of course. > > The issue between us is whether that union will be a proper subtree of > the complete infinite binary tree or will be the whole tree. This question can easily be decided by finding a node which is in the whole tree but not in the union. Should that be impossible, then a property of the subtree is that it is not a proper subset but the complete tree. > > Since every path in every one of the restricted infinite binary trees in > the union is eventually constant, every path in the union of all those > trees will also be eventually constant. That would be correct for a finite union of finite trees. What is correct for the finite case need not be correct for the infinite case. > > So that, for example, the union will not contain any path which > alternates between branching left and branching right. That would be correct for a finite union of finite trees. What is correct for the finite case need not be correct for the infinite case. > > Further, it will not contain any path which has infinitely many > branchings in both directions. That would be correct for a finite union of finite trees. What is correct for the finite case need not be correct for the infinite case. Regards, WM
From: stephen on 16 Jan 2007 09:35 Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: >> Cantor's diagonal can be represented as a (diagonal) path in an >> infinite matrix. This matrix is the union of finite matrices. It is the >> same as with the trees. > Excuse me for jumping in somewhere, but I've discovered another thing > with trees. Especially with the following one, the Stern-Brocot tree: > http://www.cut-the-knot.org/blue/Stern.shtml > It is remarked that, on the right side of the tree, there are only the > natural numbers: 1/1, 2/1, 3/1, 4/1, 5/1 .. n/1 . Each row in the tree > contains 2^(n-1) fractions. When we have arrived at n/1, there are 2^n > fractions in total (also counting 0/1 = 0). It is known that this tree > exhausts all fractions. It exhausts all naturals as well. At each stage > (n-th row) in the S-B tree, there are n naturals and 2^n fractions. We > conclude that n naturals correspond with 2^n fractions. But, since the > cardinality of the naturals is Aleph_0, the cardinality of the rational > numbers (= "reals") must be 2^(Aleph_0): Continuum Hypothesis proved .. > Han de Bruijn All you have proven is that you do not know what the Continuum Hypothesis is. Stephen
From: Han de Bruijn on 16 Jan 2007 09:44 stephen(a)nomail.com wrote: > Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote: > >>mueckenh(a)rz.fh-augsburg.de wrote: > >>>Cantor's diagonal can be represented as a (diagonal) path in an >>>infinite matrix. This matrix is the union of finite matrices. It is the >>>same as with the trees. > >>Excuse me for jumping in somewhere, but I've discovered another thing >>with trees. Especially with the following one, the Stern-Brocot tree: > >>http://www.cut-the-knot.org/blue/Stern.shtml > >>It is remarked that, on the right side of the tree, there are only the >>natural numbers: 1/1, 2/1, 3/1, 4/1, 5/1 .. n/1 . Each row in the tree >>contains 2^(n-1) fractions. When we have arrived at n/1, there are 2^n >>fractions in total (also counting 0/1 = 0). It is known that this tree >>exhausts all fractions. It exhausts all naturals as well. At each stage >>(n-th row) in the S-B tree, there are n naturals and 2^n fractions. We >>conclude that n naturals correspond with 2^n fractions. But, since the >>cardinality of the naturals is Aleph_0, the cardinality of the rational >>numbers (= "reals") must be 2^(Aleph_0): Continuum Hypothesis proved .. > > All you have proven is that you do not know what the Continuum Hypothesis is. LOL! Should I add a dozen smileys ? :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) Han de Bruijn
From: William Hughes on 16 Jan 2007 09:46
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > > > > > > > > >> Recall: Do get it out you must have put it in. > > > > > > > > > > Recall Cantor's diagonal. It consists only of finite initial segments. > > > > > Either it does not exist - or there are infinite paths in the union of > > > > > finite trees. > > > > > > > > False dilemma. It "consists" of only finite initial segments (not > > > > finitely many therof, of course) _and_ it does exist _and_ there are > > > > only _finite_ paths _as_ _members_ in the union of all finite trees. > > > > > > > > > The union of all finite trees and the complete infinite tree are > > > identical with respect to nodes and edges, but note with respect to > > > paths. Could you please specify the asserted difference concerning the > > > paths p in terms of their definition? The paths are: > > > p = Sequence (a_n) with n in N and a_n in {0, 1} > > > while the real numbers are represented by: > > > r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. > > > > > > > > > > > > > > Again: To get it out you must have put it in. Since noone put infinitely > > > > deep trees into the union of trees it does not have such paths as > > > > members. > > > > > > But as limits. I cannot believe that countable many paths can have > > > uncountably many limits. > > > > Why not? A limit of finite paths is not defined by a single finite > > path > > but by a sequence of finite paths. > > So we need even *more* (than one) objects of the tree to define *one* > limit? This makes *more* limits than objects exist? Yes. Because each object can be part of more than one limit. (Indeed each object is a part of an uncountable number of limits) > > > There are certainly a lot more sequences of paths than there are paths. > > Every sequence *is* a path. Therefore there cannot be more sequences > than paths. There are more (potentially) infinite sequences than finite paths. A (potentially) infinite sequence is not a finite sequence. A (potentially) infinite path is not a finite path. A (potentially) infinite sequence (or path), is the limit of a (potentially) infinite sequence of finite sequences (or paths). - William Hughes |