Cantor Confusion
in [Math]
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From: Virgil on 3 Jun 2007 19:22 In article <1180903197.893443.23620(a)g4g2000hsf.googlegroups.com>, WM <mueckenh(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.
From: Virgil on 3 Jun 2007 19:34 In article <1180903562.459979.290090(a)q69g2000hsb.googlegroups.com>, WM <mueckenh(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. .... > > 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. But that is only the case ini finite trees, and clearly is false in infinite trees. In fact, to "separate" any one path from ALL other paths in our CIBT requires an infinite subset of its nodes, and no finite subset suffices. Consider, for example, the path always branching left. Since for any finite set of nodes on that path there is another path that does not branch right until AFTER passing through every node of that finite set, and a few more besides, that finite set cannnot separate those two paths. Thus WM is arguing from false premises. Again! WM is perhaps confused by his MathUnRealism inspired expectation that all properties of finite trees must naturally govern infinite trees.
From: Dik T. Winter on 3 Jun 2007 21:23 In article <1180903197.893443.23620(a)g4g2000hsf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 26 Mai, 03:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179864527.794567.66...(a)p77g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > > But from > > > > your book: > > > > "also eine sogenannte normale irrationale Zahl, die keine erkennbares > > > > Muster der Ziffernfolge aufweist" > > > > > > Note the last sentence! > > > > 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. > > > > In the first place, a normal number can have a well defined sequence of > > > > digits (as the Champerowne numbers show). > > > > > > Of course, but then we could compute the numerals. > > > > Yes, we could indeed. > > Therefore I stated the remainder! Yes, and you should have left out the first part. > > > > And, in the second place, there > > > > are numbers without a well defined sequence of digits that are *not* > > > > normal. > > > > > > And there are red cars, which are not normal cars. > > > > > > > (Normal here meaning normal to a particular base.) The two are not > > > > equivalent. So even if no rule can be given for the digits of pi, that > > > > does *not* mean that it is normal. > > > > > > But if pi is normal with no recognizable pattern, then no rule can be > > > given. (If I say A ==> B then I do not imply B ==> A.) > > > > Normality of pi has nothing to do with it. > > If pi is normal with no recognizable pattern, then no rule can be > given. Right. > > If pi is not normal and with no > > recognizable pattern, then also no rule can be given. > > But I said: "If pi is normal with no recognizable pattern, then no > rule can be given." And this statement is true. I used a normal number > as an example. I did not mention the reversed statement, in particular > I did not state that normality is necessary. Can't you follow a logic > conclusion any longer? 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? > > To wit: > > normal => no recognizable pattern > > and > > no recognizable pattern => normal > > are *both* false. > > 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. 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. -- 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 3 Jun 2007 21:30 In article <1180903562.459979.290090(a)q69g2000hsb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 26 Mai, 04:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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. Oh. Above I wrote what we know about it. In mathematics both P and P' are natural numbers. > > > > You are obviously confused. For an injection from paths to nodes > > > > you need a catalogue of all paths. > > > > > > The tree *is* a catalogue of path, namely of *all* existing paths > > > which represent existing real numbers of the interval [0, 1]. > > > > 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. > > No. 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. 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. That is false. The truth is that for each two paths there is a particular node where they separate. > > > 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? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 4 Jun 2007 10:19
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: > > On 26 Mai, 04:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > 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. > > 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. > > > > > > You are obviously confused. For an injection from paths to nodes > > > > > you need a catalogue of all paths. > > > > > > > > The tree *is* a catalogue of path, namely of *all* existing paths > > > > which represent existing real numbers of the interval [0, 1]. > > > > > > 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. Remember the set of constructible numbers. There is no bijection with N - there is no list. I need merely the simple fact that there can be no more separated paths than separations in the tree. > > > > No. 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. > > 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. > 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? You consider ghost paths, which do not 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. Separations happen by nodes. Every node generates exactly one separation. > > > > > 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. 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. 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. Regards, WM |