Cantor Confusion
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From: mueckenh on 21 Dec 2006 14:19 Dik T. Winter schrieb: > > > But, indeed, if you pay for it, anything can be published. But > > > that was already true in Galilei's time. > > > > You are in error. > > It is certainly true, also at that time there were publishers enough that > would publish anything. And indeed, in 1637 a book by Galilei was > published in Leiden showing his views: "Discorsi, e dimostrazioni > mathematiche, intorno à due nuoue szienze". And I do not think that > Galilei even paid for that publication. The manuscript was smuggled > from his house (where he was in detention) to the publisher. Correct. Why had it to be smuggled? Because hardliners wanted to maintain their false worldview! In 1624 Pope Urban VIII (former Cardinal Barberini) had allowed the printing of the dialogue about the two word systems. No smuggling was necessary. > And, > yes, in the low countries almost anything could be published. Many > important books have been published in those times in the Netherlands > (while it was still in war with Spain). In the first republic ever in > the western world, there was a lot of tolerance. This role has now been taken by institutions like Shaker Verlag. > Note: my institution is closed from 25 December > to 1 January. > -- Merry Christmas. Regards, WM
From: Virgil on 21 Dec 2006 14:31 In article <1166702365.877266.8470(a)48g2000cwx.googlegroups.com>, "Albrecht" <albstorz(a)gmx.de> wrote: > On 20 Dez., 14:55, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > Albrecht wrote: > > > There is not set N. > > Should say: There is no set N. > (Is your English as weak as you are unable to interpret the wrong > spelling?) It is not necessarily a misspelling, it could be merely an unfamiliarity with English usage. "There is not a set N." correctly expresses "Albrech"'s faith.
From: Virgil on 21 Dec 2006 15:05 In article <1166728035.422113.121060(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > remember that N contains only finite numbers. > > > But infinitely many of them. > > And even more paths in the tree. We know your creed. Nevertheless it is > wrong. My "creed", which is quite verifiable, is that in ZFC or NBG, every infinite binary tree has countably many edges and nodes but uncountably many paths. WM's attempts to counter it all require assumptions provably false in both ZCFC and NBG, and not provable anywhere without assumptions provably false in ZFC and NBG. > > Consider the model given within any infinite tree by any of > > its (infinite) paths. > > > In this model, every natural appears in some line. > > > In this model, there is no line in which every natural appears. > > > This model disproves WM's contentions about his EIT. > > This model has been devised in order to show that, *given, aleph0 > exists*, 2^aleph0 > aleph0 can be contradicted. If aleph_0 represents the cardinality of N and 2^aleph_0 represents the cardinality of the set of functions from N to {0,1} then, at least in ZFN and NBG, 2^Aleph_0 > aleph_0 cannot be contradicted. > > > > Show me two elements wich cannot belong to the same line. > > > How is that in any way relevant? What I have said allows that for > > each pair there is SOME line which they do not both belong to. > > How is that in any way relevant? What I have said allows that for > each pair there is SOME line which they do both belong to. That in no way prevents the diagonal from being infinite while every line is finite. > > Are you really thinking that your statement is in any way stronger than > my? It is certainly more relevant to the issue of whether there can be a diagonal which is not contained in any line. And when the EIT is modeled in ZFC or NBG, the diagonal cannot be contained in any line but every line will be contained in the diagonal. > > > > >> Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > >> different paths and E(n) = 2*2^n - 2 different edges. So we find > >> lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, > >> E(n)/P(n) >= 1. Nevertheless "E(n)/P(n) < 1 in the actual infinity". > > > Why not? There is no reason for "continuity" to be required here. > > But in the bijection > 1, 2, 3, ... > 2, 4, 6, ... > there is some reason for continuity and some proof of same cardinality? By the very definition of cardinality, existence of any bijection between two sets establishes equality of cardinality. But the issue of "continuity", in the absence of any topology, is irrelevant. > > > By the same limit argument as WM has just claimed, there must be as > > many leaf nodes in the infinite case as there are paths, since the > > ratio of number of leaf nodes to number of paths is identically 1 for > > all finite trees. > > No, there are no leaf nodes in my tree. Then the limit of number of leaves/number of paths as n --> oo need not be 1, even though it is 1 for all finite n? In that case, neither need WM's limit claim hold. > But for any path which > separates itself from another path there is an edge where this happens. > This cannot be circumvented by the magic of infinity. Therefore there > are not more paths than edges possible. Only in WM's finite universes. We have seen that the limit of what happens in finite trees need not dictate what is the case in infinite trees (or infinite paths would all have terminal nodes) so that WM's argument fails. > > > > >> We are talking about shares, but we do not need more than 3/2 per > >> path. > > > > > How are the edges at the root node to be divided up into shares, and to > > which paths are those shared to be assigned? > > > Then the same questions need answering about the next level of edges, > > and then the next, and so on ad infinitum. > > Didn't you say that there is no continuity in infinity? What has continuity to do with anything? > > > >> If there are infinitely many different equations of form > >> "1+1+1+...+1 = n, > >> then there must be infinitely many different sums of ones. > >> Actual > >> infinity of the umber of equations implies actual infinity o at least > >> one sum of ones. > > > > > That is only true in WM's mythical axiom system, not in any actaul axiom > > system. > > If infinity is to be only potential That also is only true in WM's mythical axiom system, not in any real axiom system. If infinity is to be actual, then we have a number greater than > any natural number for the lines and simultaneously for the columns of > the EIT. But that "number", if it is to be called a number at all, is not a natural number, as are the line numbers, and line lengths.
From: mueckenh on 21 Dec 2006 16:38 Dik T. Winter schrieb: > In article <1166617425.221440.15570(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > For the evaluation of this *sum* we need *no limit* but only the > > > > computation > > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > > > We were talking about *shares*. In how many shares is the edge going left > > > from the root divided, and how do you do that division? Pray keep to the > > > subject. > > > > We are talking about shares, but we do not need more than 3/2 per > > path. > > 3/2 of what? But again you do not answer my question. In how many shares is > the edge going left from the root divided, and how do you do that division? It is irrelevant for my proof. It is as irrelevant (and silly) as the question: in how many shares is the unit divided in the last term of the series 1/2^n. Nevertheless we can calculate the limit of this series. > > > > > there is no *definable* well-ordering). So I would prefer you > > > > demonstrate it. > > > > > > That is similar to asking me to prove the parallel axiom without the > > > parallel axiom as basis. > > > > No, Dik. That is an error, perhaps it is that one whole set theory > > rests on. I did not ask you to prove the well ordering of the reals. > > That has already been done by Zermelo. I asked for a well ordering, in > > form of an explicit formula, or a recursion or a list or what ever you > > will present to enable me to find out which real number follows on > > which real number. That cannot be done. Therefore I doubt that the > > proof has any relevance. > > Indeed, that can not be done, until at some moment there is a proof > (without AC) that that axiom is true. So until such a moment we have > to do with two forms of mathematics, one with AC and one without it. > But when we assume AC, a well-ordering can be constructed, as I did > show. No, you did no show it yet. You referred to some Hamel basis. A construction either says: This is the first element and that is the second etc. or it gives a prescription how this can be found. > > But apparently you have problems when something can be shown to exist > (in the mathematical sense) but can not be computed. That is close > to the intuitionistic view. I stated that there is no possible way to construct a well-ordering. You opposed. You cannot sustain your statement. That's all. > > > But I explicitly stated how you could construct > > > a well-ordering when AC is true. You can not do it in general, and > > > nobody can, because AC is not necessarily true. > > > > Are there axioms which are necessarily true? Not according to your > > point of view. > > Indeed. No axiom is necessarily true. How then should the truth of AC be proved? Regards, WM
From: mueckenh on 21 Dec 2006 16:40
Dik T. Winter schrieb: > In article <1166617872.309270.165750(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > The inequality holds for all finite paths. Why it also should hold for > > > infinite paths escapes me. > > [ This was about the number of edges in finite case n: 2^(n+1) - 2, > and the number of paths in finite case n: 2^n.] > > > > Because we can prove that 2 - 1/2^n is always, i.e., for all finite > > natural n, less than 100. > > Yes, so I did state. > > > And because even infinity does not supply > > other than finite natural numbers. > > That has no significance to the question. You fail to see that in > the finite case each path has a last edge, but in the infinite case > *no* path has a last edge. The sum 1 + 1/2 + 1/4 + ... has no last term. Nevertheless we know, even in ZFC, that it is larger than 1 and less than 10. That has significance to our question. The series exactly maps the infinity of the paths. Regards, WM |