Another AC anomaly?
in [Logic]
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From: Virgil on 19 Dec 2009 15:14 In article <2fac8bb1-4c90-4421-b559-1ea7f0301d4f(a)e27g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 18 Dez., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > �> > Aha, you are clearly a mindreader. �Well, as far as I know mindreading > > �> > is not part of mathematics. �Anyhow, I can think of numbers larger > > than > > �> > that path. > > �> > But that is completely irrelevant. �I am able to think about a set > > that > > �> > contains all natural numbers, you apparently are not. > > �> > > �> How do you know that without confirming it by thinking the last too? > > > > Why need I to think about a last one (which there isn't) to be able to > > think > > about a set that contains all natural numbers? �Apparently you have some > > knowledge about how my mind works that I do not have. > > Yes. A very convincing and often required proof of completenes of a > linear set is to know the last element. Knowing the allegedly last element in a set does not work unless one has imposed a linear ordering on the set, which is in no wise necessary for sethood. Which is the "last" point of the set of points on a circle? > One of the rebuttals has meanwhile been changed. Peter Webb > recognized: It is true that you cannot show pi as a finite decimal, > but you can't show 1/3 as a finite decimal either. > > Just what I said. > > > > �> > �> The tree contains all paths that can be constructed by nodes, using > > �> > �> the axiom of infinity. Which one would be missing? > > �> > > > �> > The infinite paths because you stated a priori that your tree did not > > �> > contain infinite paths. �So it is impossible to construct in your tree > > �> > infinite paths by the axiom of infinity. > > �> > > �> The axiom of infinity establishes the set N from finite numbers. > > > > It establishes the *existence* of a set N of finite numbers. > > What else should be established? > > > > �> It establishes the infinite paths as well in my tree from finite > > �> paths. > > > > No. �That is impossible because you stated that the paths were finite. > > What it *does* establish is the extistence of a set P of finite paths. > > It is rather silly to argue about the uncountability of the set of > paths. Then stop doing it. There is a perfectly adequate proof that one cannot have a list of paths that contains all paths, or , equivalently, one cannot have a list of all subset of N. Since being capable of being listed is a necessary requirement for countability, that proof eliminates countability for the set of all paths s well as for the power set of any infinite set. > Only minds completely disformed by set theory could try to > defend the obviously false position that there were uncountably many > paths. Minds thus "deformed" by set theory are often capable of creating valid proofs that WM not only cannot create but also cannot even follow. > > But "10 Questions" will give you the answer why there are not > uncountably many paths. There are no infinite decimal expansions of > real numbers. There are not due paths in the tree. 1 = 1.000... is a real number with an infinite decimal expansion, and in ZF there are infinitely many more of them. I know not what a "due path" is, but in my infinite binary tree, whose nodes are the members of N, every path is an infinite subset of N, and there are more of them than even WM can count. > > It is as impossible to express any real number by an infinite decimal > expansion as it is impossible to express 0 by a unit fraction. The > rest will be explained in "10 Questions". Therefore I will stop with > this topic here. Then in WM's world, 1.000... is not a real number. Odd of him, but he has always been odd.
From: Quaestor on 19 Dec 2009 15:24 In article <9ad63bf1-549b-4822-bf86-839dd7c64d58(a)j4g2000yqe.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 18 Dez., 15:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article > > <2cd3c08a-a072-4ef5-bc0b-f0aaa9126...(a)a21g2000yqc.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > �> On 17 Dez., 15:08, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > �> > In article > > <6a57309a-a136-430c-a718-e38518c65...(a)q16g2000vbc.googlegroups= > > �> .com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > �> > �> On 16 Dez., 03:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > �> > �> > �> ({1} U {1, 2} U {1, 2, 3} U ... = ({1} U {1, 2} U {1, 2, 3} U > > �> > �> > �> U {1, 2, 3, ...} > > �> > �> > > �> > �> That is a matter of taste. > > �> > > > �> > No, it is a matter of convention. �In mathematics > > �> > � � a, b, c, ..., z > > �> > means a, b, c, continue this way until you reach z. �But starting > > �> > � � {1}, {1, 2}, {1, 2, 3} > > �> > and going on you never reach > > �> > � � {1, 2, 3, ...} > > �> > > �> That is true. Therefore it does not exist. > > > > That you can not get there step by step does not mean that it does not > > exist. > > That you cannot get step by step to 1/0 does not mean that it does not > exist? That's what he said! That's what he said! He said that! > > > > �> � � � � � � � � � � � � � � � � � � � � � �However, see Cantor, > > �> collected works, p 445: > > �> 0, 1, 2, 3, ... w_0, w_0 + 1, ..., gamma, ..., > > �> He seems to reach far more. > > > > Right, he uses a convention that is no longer used. > > Wrong, it is used presently, for instance by myself. > > > > �> > �> > A sequence of paths is not a path. > > �> > �> > > �> > �> But a union of paths is. > > �> > > > �> > No. �Suppose we have the paths 0.000 and 0.100, what is their union? > > And > > �> > is it a path? > > �> > > �> No. But the union contains two paths. > > > > Wrong. �If we look at the paths as sets, they are sets of nodes. �Their > > union is a set of nodes, not a set of paths. �And as a set of nodes we > > can form from them seven different paths. > > Wrong. The nodes of two paths give exactly two paths. But as I already > said, it is so silly to argue about the uncountability of the paths in > the tree that I will stop it here. Then WM has been silly for years. And will no doubt continue being silly. At least as long as he fails to comprehend that Cantor's several proofs of uncountability of some sets trumps WM's lack of proofs to the contrary. > > �> Let every finite path of every infinite path be mapped on the elements > > �> of omega. That was simple. > > > > By your statements infinite paths do not exist. �But pray give such a > > mapping. �Until now you have only asserted that such a mapping exists > > without showing that. > > Do you accept the mapping from omega on SUM{k = 1 to n} 3*10^-k yields > the infinite decimal expansion of 1/3? It merely creates an infinite sequence of rationals or finite decimals numerals whose numerical values converge to the number whose rational numeral is "1/3".
From: Virgil on 19 Dec 2009 15:34 In article <8baa45a8-2ba4-464b-9598-5656c43a7456(a)j19g2000yqk.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 18 Dez., 15:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article > > <dd852490-e622-4bc4-b858-dfcc3b142...(a)l13g2000yqb.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > �> On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: > > ... > > �> > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > �> > > > No. �Suppose we have the paths 0.000 and 0.100, what is their > > union? > > �> > > > And is it a path? > > �> > > > �> > > No. But the union contains two paths. And an infinite union of that > > �> > > kind may contain the path 0.111... > > �> > > > �> > So according to WM a union of sets may contain an object not contained > > �> > in any of the sets being unioned. > > �> > > �> For paths and initial segments to contain and to be is the same. > > > > Eh? �paths contain nodes and initial segments contain numbers. > > But be aware that you use the word 'contains' with two different meanings > > at different times. > > For paths we have: Paths contain initial segments and paths are > initial segments. Wrong! An intial segment having a last node cannot be a path if its last nose is not a leaf node. In this sense, an initial segment that is not maximal in a given tree is not a path in THAT tree. > > > > �> For paths and initial segments to contain and to be is the same. In > > �> fact: > > �> {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > > �> and also > > �> {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. The second of those alleged equations is nonsense as expressed, as {1, 2, 3, ...} is not the successor to any member of the series {1}, {1,2}, {1,2,3}, ... In order for it to make sense it would have to be rewritten as ({1} U {1, 2} U {1, 2, 3} U ... ) U {1, 2, 3, ...} = {1, 2, 3, ...} so the the first ellipsis and the second do not overlap. > > > > In the mathematical sense the union contains numbers, not sets. > > The union contains subsets. Not as members. The members of the st being unioned are subsets of the union, but
From: WM on 19 Dec 2009 16:36 On 19 Dez., 21:14, Virgil <Vir...(a)home.esc> wrote: > > Yes. A very convincing and often required proof of completenes of a > > linear set is to know the last element. > > Knowing the allegedly last element in a set does not work unless one has > imposed a linear ordering on the set, which is in no wise necessary for > sethood. Which is the "last" point of the set of points on a circle? I am interested in linear sets. > > It is rather silly to argue about the uncountability of the set of > > paths. > > Then stop doing it. There is a perfectly adequate proof that one cannot > have a list of paths that contains all paths, or , equivalently, one > cannot have a list of all subset of N. You are wrong. The "perfectly adequate proof" is wrong. > > Only minds completely disformed by set theory could try to > > defend the obviously false position that there were uncountably many > > paths. > > Minds thus "deformed" by set theory are often capable of creating valid > proofs that WM not only cannot create but also cannot even follow. There is one simple proof: Every node of the binary tree splits one path into two paths. the number of paths cannot surpass the number of nodes. Nobody would ever have doubted that without the unfortunate development of mathematics induced by Cantor. > > > > > But "10 Questions" will give you the answer why there are not > > uncountably many paths. There are no infinite decimal expansions of > > real numbers. There are not due paths in the tree. > > 1 = 1.000... is a real number with an infinite decimal expansion, and in > ZF there are infinitely many more of them. From the sequence of digits you cannot obtain 1. First you must be sure that there never will appear a digit other than 0. But unless you know the last digit, you cannot be sure. And you cannot know the last digit. Of course you can know a formula saying digit d_n = 0 for all n in N. But that is a finite formula. It is not an infinite sequence of digits that informes you about the due number. Regards, WM
From: Marshall on 19 Dec 2009 17:17
On Dec 19, 1:36 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > From the sequence of digits you cannot obtain 1. First you must be > sure that there never will appear a digit other than 0. But unless you > know the last digit, you cannot be sure. And you cannot know the last > digit. > > Of course you can know a formula saying digit d_n = 0 for all n in N. > But that is a finite formula. It is not an infinite sequence of digits > that informes you about the due number. I think that may be your lamest argument yet. Marshall |