Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Dik T. Winter on 20 Dec 2006 19:59 In article <1166616863.359185.111540(a)n67g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Right. That is the difference between scientific press and vanity press. > > In the former there is review and editing assistence, in the latter you > > can publish anything you want to publish. In the former you do not pay > > anything, > > Never heard of "page charge"? I remember that was usual with Physical > Review already 30 years ago. No. And I now see that it is usual in physics and astromical journal. As far as I know it is not usual at all in mathematical journal. And I just checked the AMS website, and there is no mention of it at all on their website, except for some discussion in 1977, but apparently never implemented. Also I never have heard any mention about it when I did (co-)author some articles. As far as I know in the mathematical community there are no charges for publishing, whether as a paper in a journal or a book. And I think that many institutions would have to stop publishing if the kind of money is asked that some physics papers do. > > 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 � 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. 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. The only reason you are in a better position is that you do not have to smuggle the manuscript out of your house. > > But whatever, send me a copy of the book if you want a review in this > > newsgroup. I promise to donate it to our library when finished. > > I'll do it. Well, I wait and I'll see. Note: my institution is closed from 25 December to 1 January. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: cbrown on 20 Dec 2006 20:18 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > > No. But I need no proof in order to see that > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > > > > Yes, it appears that you need no proof to make any of your assertions. > > You simply "see" them; or "feel" them. That's not what I would call > > "really correct mathematics". > > Call it whatever you want. I call it "mathiness". > I can calculate 1 + 1/2. Good for you! > But considering the > fact that you would call the following result "correct mathematics", I > prefer to stay with my version. > > 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". It is /not/ a fact that I would call this "result" correct mathematics. I would call it yet another example of your devotion to "mathiness" whenever you claim to provide a "contradiction". While "E(n)/P(n)" is certainly well-defined when both E(n) and P(n) are real numbers, it is not well-defined otherwise. You simply "feel" that Cantor's result must imply "E(n)/P(n) < 1 in the actual infinity"; without considering what that assertion /means/. You apply "mathiness" to the problem. You "know" (need no proof) that "2 < 3" implies that "2/3 < 1". You "see" the assertion "|N| < |R|", and because it "looks" like "2 < 3", you "feel" that you can therefore conclude "|N|/|R| < 1". Instead, the statements "|N|/|R| < 1" and "E(n)/P(n) < 1 in the actual infinity" are simply mathematical nonsense: neither true nor false. Equally, if we "stay with your version", "E(n)/P(n) >= 1 in the actual infinity" is nonsense; just as "Triangle/Square <= 2" is nonsense. On the other hand, "down to level n", E(n) is finite for all finite n. Nevertheless, "E(n) is not finite in the actual infinity". Do you consider the above also not "correct mathematics"? What is "incorrect" about it? Cheers - Chas
From: Dik T. Winter on 20 Dec 2006 20:10 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? > > > 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. 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. > > 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. > > If somebody came up > > with a construced well-ordering of the reals that would imply that AC > > is true. In a similar way, if somebody could construct a line parallel > > to another and show that there are no other parallel lines, that would > > prove that the parallel axiom is true. > > The parallel axiom is true on a sheet of paper. Where is AC true? The only thing I can show on a sheet of paper is that given a line on that sheet and a point, I can draw several lines through that point that do not meet the given line, and so are parallel to the given line. You can show it true only on an actually infinite sheet of paper. You know that you can do elliptical geometry on an elliptical sheet of paper (that is one of the models of elliptical geometry)? -- 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 20 Dec 2006 20:22 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. (Assuming ZF or something similar of course, in which we are reasoning, I think.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Newberry on 20 Dec 2006 21:43
mueckenh(a)rz.fh-augsburg.de wrote: > Newberry schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Newberry schrieb: > > > > > > > > > > >The edges have cardinal number aleph_0. > > > > > > > > Certainly. > > > > > > > > > A > > > > > subset of the paths can be shown to be in bijection with the real > > > > > numbers of the interval [0,1]. > > > > > > > > Which subset? Path as a set of intervals or path as a set of eges? How > > > > does the above follow from this? > > > > > > The paths are binary representations of all real numbers of the > > > interval [0,1]. Some rational reals have double representations, namely > > > such ending by 000... and such ending by 111... like 1.000... and > > > 0.111... . So there are not less real numbers in the interval [0, 1] > > > than paths in the binary tree. We have > > > > > > aleph_0 = |{edges}| >= |{paths}| >= |{reals of [0,1]}| = 2^aleph_0 > > > > > > while according to set theory > > > > > > aleph_0 < 2^aleph_0 > > > > > > Regards, WM > > > > Well, a Cantorist would say that this part does not hold > > > > |{edges}| >= |{paths}| > > > > because you considered only the finite paths but not the infinite paths. > > In the infinite tree there are no finite paths. Yes, but you arrived at |{edges}| >= |{paths}| by calculating lim{n-->oo} (2*2^n - 2)/2^n = 2 In this limit all the n are finite. So a Cantorist would claim that you examined only the finite sub-paths, and not the infinite paths. I do not know if this Cantorist claim makes sense, but the question is whether you can generate a direct contradiction of the type P & ~P in ZFC. > > But if we calculate the limit of a finite tree, the result is the same. > Meanwhile we have found out that set theorists must assume 1 + 1/2 + > 1/4 + ... > 10 be possible unless we have some special definition of > limits. I think his shows the whole insanity of their position well > enough. > > Regards, WM |