Cantor Confusion
in [Math]
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From: Virgil on 4 Jun 2007 16:11 In article <1180981938.556670.77030(a)q75g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 4 Jun., 01:17, Virgil <vir...(a)comcast.net> wrote: > > In article <1180902384.062112.256...(a)m36g2000hse.googlegroups.com>, > > Corrected: > > Would you also reject the possibility that the initial segments of the > diagonal > (including the complete segment) can be bijected with themselves? > > If not, why is the bijection of the initial sements of the diagonal > (including the complete diagonal) with the lines possible, but the > bijection of the initial sements of the diagonal (including the > complete diagonal) with the columns is not? Who says it is not? Unless WM is claiming that one infinite set has a greater cardinality than the other, bijection is inevitable. And there is a trivial such bijection that WM has overlooked.
From: Virgil on 4 Jun 2007 16:14 In article <1180982446.483386.3830(a)k79g2000hse.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 4 Jun., 01:22, Virgil <vir...(a)comcast.net> wrote: > > In article <1180903197.893443.23...(a)g4g2000hsf.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > If pi is normal with no recognizable pattern, then no rule can be > > > given. > > > > False! It is just that the "rule" is not simple, but one can always say > > what it takes work out the "next digit" of pi once one has any given > > number of digits. > > There is a (relatively small natural) number n which needs the > complete memory of the universe. Dealing with such practical problems is engineering, not mathematics.
From: Virgil on 4 Jun 2007 16:22 In article <1180982668.026854.204170(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 4 Jun., 01:34, Virgil <vir...(a)comcast.net> wrote: > > In article <1180903562.459979.290...(a)q69g2000hsb.googlegroups.com>, > > > > > > > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 26 Mai, 04:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1179931587.894431.146...(a)u30g2000hsc.googlegroups.com> WM > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > > On 22 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > In article <1179747854.168331.52...(a)x35g2000prf.googlegroups.com> > > > > > > WM > > > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > > > > On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > In article > > > > > > > > <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM > > > > > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > > > > > > Its is more. You cannot answer the question whether the > > > > > > > > > numbers P = > > > > > > > > > [pi*10^10^100] and P' = P with the last digit replaced by > > > > > > > > > 3 > > > > > > > > > nsatisfy P < P'. > > > > > > > > > > Yes, so what? Your distinction is just terminology, and not > > > > > > > > more > > > > > > > > than > > > > > > > > that. > > > > > > > > > It is by far more than a difference in terminology if one will > > > > > > > never > > > > > > > or always be able to answer a question. > > > > > > > > Makes absolutely no sense. You have some implicit notion of > > > > > > number in > > > > > > mind > > > > > > that I do not have. > > > > > > > Trichotomy. > > > > > > But there is trichotomy. We know that either P < P' or that P = P' or > > > > that > > > > P > P'. But we do not have the means to determine which of the three > > > > holds. > > > > > We know nothing about that number. > > > > Speak only for yourself, WM. > > If you know more than me about this number, please don't hesitate to > publish it. > > > > ... > > > > > > Each node splits a set of paths in two sets of paths. When you > > > > come in at a node with a set of K paths, you come out with two sets of > > > > K/2 paths (this is a bit informal). If K is infinite, so is K/2. And > > > > if K is uncountable, so is K/2. So all this splitting does show > > > > nothing. > > > > And it is not the case that each node separates one more path. If that > > > > were the case you should be able to indicate what path is separated by > > > > the root node. > > > > > Every path which can be identified must be separated from all other > > > paths. For this sake there must be as many nodes as separated paths. > > > > That would be the case provided for each path there was a SINGLE node > > separating it from all other paths. > > No. It is true because for every other path q which you copare with p > there must be at least a single node. That implies, falsely, that for every such node there is only one such other path, whereas there are as many other paths from each node as there are paths in the whole infinite tree. > Brefly: Every separated path comes into being only by a noded > Therefore there cannot be more separated paths than nodes. Assumes, falsely as usual, that each node of a path "separates" that path from only one other path (or even only a finite number of other paths). The truth is that in a CIBT, each node of a pat "separates" that path from uncountably many others, as many as in the entire tree. That this is easy to see is true. That WM chooses not to see it is also true.
From: Dik T. Winter on 4 Jun 2007 22:42 In article <1180966798.913696.131020(a)n4g2000hsb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 4 Jun., 03:30, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1180903562.459979.290...(a)q69g2000hsb.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: .... > > > We know nothing about that number. > > > > Oh. Above I wrote what we know about it. In mathematics both P and P' > > are natural numbers. > > Mathematicians seem to be very short sighted. If they see a great many > of numbers, they quickly assume that here are infinitely many. It is > simply a ridiculous claim. There are not more than 10^100 numbers, and > there are many natural numbers less that 10^10^100 which do not exist > and by no means can be brought to existence. That is your claim. But the axiom of infinity tells different. Each and every time when you claim this you state that you are doing mathematics without the axiom of infinity. We are doing mathematics *with* the axiom of infinity. So your claims are not relevant for the way we do mathematics. On the other hand, if there are no more than 10^100 numbers, the theorem that pi(x) and Li(x) cross becomes ridiculous. And the theorem that there are infinitely many primes. How many primes are there? Your idea of "bringing into existance" is very strange. You simply want a decimal representation, and that is it. If such does not exist you state that the number does not exist. Well, that is your way doing mathematics, but it would help you if you did build the remainder of mathematics based on your reasoning. > > > > And I thought that a catalogue in your sense numbered the paths from 1 > > > > onwards. Apparently you used a different meaning, again. > > > > > > A catalogue is not a list. "Catalogue" is not normed as far as I know. > > > It is simply a collection o all paths. > > > > So you need only a collection for an injection? Strange reasoning. But > > it was *you* who introduced the word catalogue in the discussion. If you > > keep on introducing non-normed words you reap only confusion. > > I need not a bijection in order to show that a set is countable. Indeed. But you do not *give* a bijection between nodes and paths. You give only an injection from nodes to paths. You have to show that either (1) that injection is also a surjection or (2) there is an injection from paths to nodes. But you do none. You are simply wrigling around and around. > Remember the set of constructible numbers. There is no bijection with > N - there is no list. There is a list, but the list is not constructible. There is a list because there is an injection from N to the constructible numbers, and also an injection from the constructible numbers to N. > I need merely the simple fact that there can be > no more separated paths than separations in the tree. You need a proof of that fact. > > > Every path which can be identified must be separated from all other > > > paths. For this sake there must be as many nodes as separated paths. > > > > The first is true. For the second you provide no proof, because it > > assumes that for each path there is a particular node where it separates > > from all other paths. > > Wrong. For every path there is a node where it separates from all > paths considered up to a certain level. What are "paths considered up to a certain level"? Do you mean "finite paths"? > > > That is false. > > That means that there must be as many nodes as paths for every finite > segment of the tree, but not for the whole tree? Every finite segment of the tree contains only finite paths, there are no infinite paths in such segments. And as I have already stated again and again, the number of finite paths in the infinite tree is countable because there is a trivial bijection between the finite paths and the nodes. Pair each finite path with the node where it terminates. The difference with the infinite paths is that they do not terminate. But for some reason you appear to have difficulty with comprehending non-terminating paths. > You consider ghost > paths, which do not exist. By what meaning of "exist"? > > The truth is that for each two > > paths there is a particular node where they separate. > > It is useless to follow this fancy discussion. Fact is that there can > be no more separated paths than nodes. A fact is not a fact until it has been proven. > Separations happen by nodes. > Every node generates exactly one separation. Yes, and every node separates two bunches of uncountably many paths. > > > > > Therefore, there must be as many separation points, or nodes, as > > > > > separated paths. > > > > > > > > Wrong. > > > > > > Absolutely correct. But I you wilfully adhere to some believe in ghost > > > paths, it is no longer useful to maintain this discussion. > > > > What ghost paths? > > Such which are separated without nodes but nevertheless exist of > nodes. Makes no sense. For each pair of paths there is a node were they separate. > Every node is the end of a finite path. There is simply nothing left > to prove the existence of the other (even uncountably many) paths. So you disallow the infinite paths. In contradiction with the axiom of infinity. But *that* means that 1/3 is not in your tree. Pray be consistent for once. > Therefore they are ghost paths. Hrbacek and Jech have seen that this > line of reasoning is nonsense. They say: "A branch in T is a > transfinite sequence b whose all proper initial segments are in T but > itself is not." (p. 220) And they are right, of course. But the > immediate consequence is that an infinite sequence of digits is also > missing in Cantor's list and, therefore, not available for his > diagonal proof. I have to look it up, but I trust that you thoroughly misinterprete their position. I suspect they define T as a tree of finite paths. In that case they are right. But in that case 1/3 is missing in the tree. -- 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 4 Jun 2007 22:58
In article <1180968036.887687.51870(a)p47g2000hsd.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 4 Jun., 03:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > Yes, I note it. And I note also that the first part of that > > > > sentence is nonsense, and that is what I do remark on. Whether > > > > or not pi is normal has *no* relationship with the remainder. > > > > > > Therefore I stated the remainder! > > > > But the first part is nonsense. > > No, it is a suggestive example. Most normal irrational numbers do not > exhibit a regular pattern. Do you disagree? I do not know. The only known examples of normal irrational numbers exhibit a regular pattern. > > > > Yes, we could indeed. > > > > > > Therefore I stated the remainder! > > > > Yes, and you should have left out the first part. > > Why? Do you think it is wrong? Do you think that usually normal > irrational numbers are like Champerowne numbers? Well, the only known examples are like those numbers. Do you know a normal irrational number that does *not* display a regular pattern? > > > If pi is normal with no recognizable pattern, then no rule can be > > > given. > > > > Right. > > That is what I said. Should your German be too bad? But that is not stated by you. Is your German so bad? > > No, you stated: "presumed that pi is really a number without rules, hence > > a so-called normal number that has no recognizable pattern in the sequence > > of digits". The part about "normal number" is nonsense. In German: > > "vorausgesetzt, da� pi wirklich eine regellose Zahl ist, also eine > > sogenannte normale irrationale Zahl, die kein erkennbares Muster > > der Ziffernfolge aufweist". > > What is the reason you mention normality here?# > > Because an unnormal number would easily obey a rule, for instance it > could show no 2,3,4,5,6,7,8, or 9 after a certain index, which would > increase the possibility of predicting a digit by chance. A normal would just as easily obey a rule, for instance it could be a Champerowne number, which would make any digit exactly predictable, and not by chance. But you were not talking about predictability but about recognisable patterns. Those two are not the same. Given a number that is normal to base-9, it has probably not a recognisable pattern. When you consider that representation as a number in base-10, the pattern remains just as unrecognisable as it was, the only thing you can say it that the digit 9 does not occur, which makes it a non-normal number. > > > I did not use either of them. I did not use "=>". I used a normal > > > irrational number (as a suggestive example) "and" I required that no > > > recognizable pattern is visible. > > > > What normal irrational number *did* you use as an example? pi? Do you > > have a proof that it is normal? If so you should publish it. > > I said: "vorausgesetzt", which means: "assumed", Yes, so you did not use a normal irrational number as example. > > How > > should I interprete your "also eine sogenannte normale irrationale Zahl"? > > This looks, to me, more like a definition of normal number than anything > > else. > > It is a definition. "Regellos" implies "normal", because otherwise > there would be a rule like "only few 1's after the digit number > 10^100" or so. You are seriously wrong. "Regellos" does *not* imply "normal". Moreover, the digits of pi are not "regellos", they are precisely defined by the definition of pi. I think you have to define your concept of "regellos". > Is that the only "error" which you mean to have discovered? Or why are > you so much chopping on it? I am chopping on it because it was a ridiculous statement that tended to insert some mathematics in a largely non-mathematical text. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |