Another AC anomaly?
in [Logic]
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From: Dik T. Winter on 22 Dec 2009 09:47 In article <b484e377-9dc2-424b-80c3-2912165f636c(a)a32g2000yqm.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 21 Dez., 14:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > That you cannot get step by step to 1/0 does not mean that it does not > > > exist? > > > > Indeed. On the projective line (that precedes Cantor by quite some time > > as far as I know) it does exist. > > But does the projective line exist? Exist in what sense? But it is a concept from projective geometry, but perhaps you think that is also nonsense? I may note that the point at infinity was developed by Kepler and Desargues in the 17th century. > > > > > 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. > > > > But you are not a mathematician. > > Do you think so? Yes, I think so and your inability to provide definitions and proper proofs shows it. > I studied mathematics and a university council > appointed me to teach mathematical lessons. Well, the same did hold for my father, but he never considered himself a mathematician, but a physicist. > > > > 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. > > > > Darn, the paths 0.000 and 0.100 contain the following nodes: > > 0.000 = {0., 0.0, 0.00, 0.000} and 0.100 = {0., 0.1, 0.10, 0.100} > > where a node is named by the path leading to it. Their union contains > > the following paths: > > 0., 0.0, 0.00, 0.000, 0.1, 0.10, 0.100 > > and I count seven. > > The nodes of two maximal paths give two maximal paths. You count > initial segments. I did not count segments, I counted paths. You stated there are only two paths in the union, I count seven paths. Or are those seven things suddenly not paths? > > > > 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? > > > > *What* mapping? Do you mean from n in omega -> SUM...? In that case > > the infinite decimal expansion of 1/3 is unmapped. > > The sum of all finite segments is not the infinite path. Interesting. You have stated that your paths are finite, so I consider the mentioning of an infinite path strange. Moreover, you are now talking about sums, not about paths. But *what* n in omega maps to the infinite decimal expansion? You have not stated that. And if you think the infinite expansion is mapped you should be able to state an element of omega that maps to that infinite path. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 22 Dec 2009 16:10 In article <Kv251u.sI(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > > The union contains subsets. > > > > > > This is so according to set theory. Of course it is rubbish. > > > > > > Well, if you want to use terminology in a different meaning than > > > standard, of course. > > > > To contain as a subset is not correct in English? > > It is correct English but misleading in a set theoretic context. A set can contain something as a subset OR contain something as a member, but these are different containment relationships, neither of which necessitates the other. WM has the bad habit of using "contain" ambiguously. It often seems that he is being purposely ambiguous about it. Proper mathematical usage requires that "contain" be used unambiguously, whenever there can be any doubt, by explicitly stating in each such instance whether it means the subset or the membership relation.
From: Ross A. Finlayson on 22 Dec 2009 20:27 On Dec 19, 5:20 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <500d09c4-6322-417e-a4d9-f9716b19e...(a)c34g2000yqn.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...(a)gmail.com> wrote: > > > > > On Dec 19, 12:14 pm, Virgil <Vir...(a)home.esc> wrote: > > > In article > > > <2fac8bb1-4c90-4421-b559-1ea7f0301...(a)e27g2000yqd.googlegroups.com>, > > > > WM <mueck...(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. > > > There you assume the consistency of the statement "Dedekind/Cauchy is > > complete" > > Not at all! > > I merely note that every rational integer is included among the reals, > in both the Dedekind construction and the Cauchy construction, and 1 is > nicely both rational and integral. Oh, so 1 = 1.000... now, eh. Good luck with that. Regards, Ross F.
From: Marshall on 22 Dec 2009 20:38 On Dec 22, 5:27 pm, "Ross A. Finlayson" <ross.finlay...(a)gmail.com> wrote: > On Dec 19, 5:20 pm, Virgil <Vir...(a)home.esc> wrote: > > > Not at all! > > > I merely note that every rational integer is included among the reals, > > in both the Dedekind construction and the Cauchy construction, and 1 is > > nicely both rational and integral. > > Oh, so 1 = 1.000... now, eh. > > Good luck with that. He won't need it. 1 the ratio is the same number as 1 the natural, is the same number as 1 the member of any other set. Unless you confuse numbers and their representations. Marshall
From: WM on 27 Dec 2009 12:57
On 22 Dez., 15:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Even a matheologian should understand that: If there is no digit at a > > finite place up to that the sequence 0.333... identifies the number > > 1/3, then there is no digit at a finite place up to that the number > > 1/3 can be identified. > > Right, there is no digit at a finite place up to that the number 1/3 can be > identified. And as there are no digits at infinite places that appears to > you to be a paradox. It is not. There is *no* finite sequence of digits > that identifies 1/3. But there is an *infinite* sequence of digits that > does so. Why not pronounce it a bit clearer? We know that every digit is the last digit of a finite initial segment of digits. There is no other digit, is it? Therefore your statement can be expressed in the following way: There is no finite initial segment that identifies 1/3. But there is an infinite finite initial segment that identifies 1/3. If you read this sentence, perhaps you get an impression of what matheology is? > > I said, if there is a sequence that identifies > > 1/3, then the identifying digits must be at finite places. > > Right, all identifying digits (there are infinitely many) are at finite > places. Therefore your statement can be expressed in the following way: There is no finite initial segment that identifies 1/3. But there is an infinite finite initial segment that identifies 1/3. > > > The union of finite initial segments cannot ield an infinite initial > > segment? > > Yes. But as you have stated that your tree contained finite paths only, > such an infinite initial segment is not (according to *your* definition) > a path. The axiom of infinity states: There is an infinite set such that with every n the follower of n belongs to the set. This is true for the binary tree. With every finite path 1, ..., n the follower 1, ..., n+1 is in the tree. The union of all those paths of course is finite, as I said, because there is no infinite path, but according to the axiom it is infinite. > > > Does the sequence of 1/3 not consist of a union of all finite > > initial segments? > > It is, but also (according to *your* definition) it is not a path. There can be no question: The union of all finite initial segments is finite. But according to the axiom, it is infinite. So we can both be well pleased. Regards, WM |