Cantor Confusion
in [Math]
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From: Virgil on 24 Apr 2007 14:20 In article <1177420888.924277.217920(a)r30g2000prh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 22 Apr., 19:07, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Apr 22, 9:08 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > On 22 Apr., 13:08, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > <snip> > > > > > > Is the following statement true of false > > > > > > There are enough nodes to separate an uncountable number > > > > of paths. > > > > > Definitively: False. > > > > Recall > > > > M: a set of separation points can separate a path from all other > > M: paths. > > > > Which of these statements is false? > > > > A path can be separated from all other paths by a set of nodes > > > > There are uncountably many sets of nodes > > False is the implication that n nodes can separate more than n+1 path > bunches (where a path is a path bunch with one element). False is the implication that such separtions are at all relevant to the issue of the cardinality of the set of paths i a CIBT. The levels of a CIBT are indexed by N, and for every different subset of N there is a path different from all other paths, that path ranching left at just the levels indexed by that subset of N and branching right elsewhere. So Card(set of paths) = Card(Power set of N) > Card(N). WM, being incapable of faulting this proof of the uncountability of the set of paths of any CIBT, always chooses always to ignore it.
From: Virgil on 24 Apr 2007 14:55 In article <1177433194.009828.202370(a)o40g2000prh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 22 Apr., 18:48, Virgil <vir...(a)comcast.net> wrote: > > In article <1177234322.313795.125...(a)l77g2000hsb.googlegroups.com>, > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 21 Apr., 22:55, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On Apr 21, 4:33 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > Each of these bunches contains an uncountable number > > > > of paths. All you show is that there are a countable number > > > > of bundles, each of which contains an uncountable number of paths. > > > > > What about Cantor's list? Does it contain the representations of paths > > > or of path bundles? > > > > Is a "path bundle" a set of paths? If so why not call it a set of paths? > > > > > If your answer is "paths", what is the difference > > > to a path bunch with one element? > > > > What is the difference between a "path bundle and a "path bunch"? > > I use bunch, Dik uses bundle. I see no difference. > > > > > > Each of these paths can be separated from all other paths by > > > > a set of nodes. There are uncountable many sets of nodes. > > > > There are enough nodes to separate uncountable many paths. > > > > > Your error lies in the fact, that one node can only once be used to > > > separate two path bunches. > > > > False! There are sets of paths which cannot be isolated from each other > > by any one node. > > Which sets do you have in mind? For example, the set of all baths that branch left at level 5 cannot be separated from the set of all paths which branch right at level 5 by any one node. > > > > > Therefore the "sets of nodes" (which are > > > nothing but paths) do not help to show the uncountability of paths. > > > You are lacking some logic (quite a lot). You try to prove the > > > uncountability of a set by the uncountability of that set. > > > > False! > > > > We succeed in proving uncountability of the set of paths of a CIBT by > > proving that the set of paths bijects with the uncountable power set of > > the set of levels of a CIBT. > > That is nonsense. These uncountability proofs are wrong. Saying that they are wrong does not make them wrong. You must show that some step in their proof is wrong in order to establish that they even can be wrong, much less that they are wrong Here it is again: (1) The set of levels in a CIBT is indexed by N. (2) For each subset of N there is a unique path which branches left at all the levels indexed by that set and right elsewhere. (3) this establishes a bijection between the set of all paths in the CIBT and the set of all subsets of N (4) Card(set of all subsets of N) = Card(P(N)) > Card(N) (5) Card(set of all paths in a CIBT) > Card(N). (6) The set of all paths in a CIBT is uncountable, meaning of cardinality greater than that of N. Which step is wrong, WM? If None are, then WM is wrong. > Further, if > they were correct, I show a contradiction in set theory. Your alleged rebuttal does not occur in any set theory recognized by mathematics , nor in any set theory which does not presume the contrary a priori, nor in any set theory all of whose presumptions (axioms) have been clearly stated so that one can determine unambiguously what does or does not follow from those axioms. > This cannot > be rebutted by a different proof. Each such attempted proof has been validly rebutted by showing one or more flaws in that proof, frequently by different critics. Whereas WM has not validly rebutted any of the several proofs of uncountability, such as the one above. Wm is in over his head.
From: William Hughes on 24 Apr 2007 16:52 On Apr 24, 9:21 am, mueck...(a)rz.fh-augsburg.de wrote: > On 22 Apr., 19:07, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On Apr 22, 9:08 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 22 Apr., 13:08, William Hughes <wpihug...(a)hotmail.com> wrote: > > > <snip> > > > > > Is the following statement true of false > > > > > There are enough nodes to separate an uncountable number > > > > of paths. > > > > Definitively: False. > > > Recall > > > M: a set of separation points can separate a path from all other > > M: paths. > > > Which of these statements is false? > > > A path can be separated from all other paths by a set of nodes > > > There are uncountably many sets of nodes <snip complete failure to address the question> Try again. Which of these statements is false? A path can be separated from all other paths by a set of nodes There are uncountably many sets of nodes -William Hughes
From: Dik T. Winter on 24 Apr 2007 22:01 In article <1177158861.675889.121570(a)y80g2000hsf.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 20 Apr., 16:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1176981294.085164.20...(a)n76g2000hsh.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 18 Apr., 16:56, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > Without the chance that uncountably many are separated in > > > > > the whole tree? > > > > > > > > They are all separated from each other. I do not know why you need > > > > uncountably many separation points for that. > > > > > > Because one node separates only one (bunch of) path(s) from another > > > one. > > > > It makes sense only if you remove the parenthesis. When you remove the > > stuff between parenthesis it makes no sens. > > That depends on the notation. The "bunch of" makesw it somewhat > clumsy. This does not make a difference. When you remove the stuff between parenthesis it makes no send. There is *no* node that separates a specific path from another one. > There are bunches of paths like the bunch 0,1 which contains all paths > from 0,1000... to 0.111... > A searated path is a bunch of paths with only one element like > 0,111... > > In order to get two separateed bunches of paths, there is one bunch > required which contains all these paths and one node to split the > incomin bunch into wo bunches. As far as I understand from your book, there is no splitting of bunches. There are only nodes where bunches terminate. According to the book a bunch of paths is a set of paths with a common initial segment (i.e. set of nodes). So at each node at least three bunches are involved (actually more). There is the bunch that terminates with the node in question, and there are the two bunches that go left and right. All three are incoming bunches of the node, and only the latter two are outgoing bunches. > The infinite tree contains all separated paths. Therefore there are as > many nodes (minus one) required in the tree. A proof is still required. > Should no separated paths exist, then they would also not exist in > Cantor's list as entries or as diagonal number. Makes no sense. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 24 Apr 2007 22:28
In article <1177159491.684039.144700(a)y80g2000hsf.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 20 Apr., 17:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1176981838.819590.80...(a)l77g2000hsb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > On 18 Apr., 16:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > The mathematical question is: "are all paths available in the infinite > > > > tree", and the answer is: "yes". A follow-up question can be "is each > > > > path separated from all other paths", and the answer is, again, "yes". > > > > And a further question can be, "is there a level such that a particular > > > > path is separated from all other paths", and the answer to this is "no". > > > > "exist" is barely a mathematical term. > > > > > > Existence of separated paths is not possible without being separated. > > > But the set of occasions for separations (= nodes) is countable. > > > > In what way is that a contradiction? As each path separates at each of > > its elements (nodes) from uncountable many other paths, I see no > > contradiction. > > The tree must contain all separated path due to real numbers between 0 > and 1. For this sake we need as many nodes as separated paths. Still no proof. > > > > What is a "point of separation"? > > > > > > A node, of course. > > > > So you are stating that there are no more paths in the tree than there are > > nodes? But *that* is obviously false, as any finite tree shows. > > Every finite tree contains less separated bunches of path than nodes. Wrong, I think. But that depends on your definition of "separated bunches of path", which you do not give. > > > > But indeed, in every finite tree with n levels there are 2^n+1 finite > > > > paths. > > > > > > No, there are 2^n finite paths. > > > > Now again, you are limiting paths to those that go on to the end. When > > you do *that*, you get problems when you consider sets of paths. > > Instead of "path" which seems to be misleading use "bunch of paths". Well, regarding the conflicting definitions of "bunch of paths" in your book, I wonder. But I still maintain that you are getting difficulties when you consider sets of paths. As according to your definition of path in trees (all go through until the end), given two finite trees T_n and T_m with n different from m, there is *no* path in T_n that is also a path in T_m, as is the case with the reverse. And as (according to your definitions) a path bundle is a set of paths, there is *no* path bundle in T_n that is also a path bundle in T_m. > > > > And in the infinite tree there are countably many finite paths. > > > > But this says *nothing* about infinite paths. > > > > > > Cantor's diagonal can be defined up to any line n of the list. But > > > this says *nothing* about an infinite sequence of digits. > > > > It is defined for *any* line of the list. > > That does not define it for *all* lines. It does. > Here is an example. (You know that all rational numbers obey the laws > of trichotomy.) > > Cantor's (first diagonalization) method allows to enumerate every > rational number. So each rational has its place in the sequence. But > this sequence does not enumerate *all rationals". Otherwise there must > be the largest rational less than 1. Wrong. That must only be true if there is a largest natural number. Consider the following mapping: Let's have 'r' a rational number > 0. Rewrite 'r' as a continued fraction [r0, r1, ..., rn+1], where the ri are natural numbers, except for r0, which can be 0. It is easily shown that such a representation is unique. Next map that number to: 2^r0.(1 + 2^r1.(1 + ... (1 + 2^rn)...)). So the continued fraction is mapped to an integer which has in its binary expansion a specific spacing of the 1 bits. From this a reverse mapping is obvious. Pray tell which rational does not map to an integer, and which integer does not map to a unique rational. And also tell me which integer maps to the largest rational smaller than 1. > The absence of this rational is a natural as he absence of the set > {N,k,f} in Hessenberg's proof. That is only because you put the cart before the horse. You do not understand the conditions. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |