Cantor Confusion
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From: cbrown on 14 Dec 2006 22:17 mueckenh(a)rz.fh-augsburg.de wrote: > The ordinal 2^omega is a countable set. Even omega^omega is a countable > set. > You could learn this elementary knowledge from my book: > http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN=3-8322-5587-7 > Unfortunately, following this link led me to a page saying "Das Buch ist nicht in unserer Datenbank gespeichert", which I don't understand, but guess means "That book is not in our database, sadly". Cheers - Chas
From: G. Frege on 14 Dec 2006 22:42 On Fri, 15 Dec 2006 03:18:01 +0100, Ralf Bader <bader(a)nefkom.net> wrote: > > M�ckenheim, however, commits a similar but much more stupid error on p.5 of > the remarkable flatulency he put on the arxive today [...] > Remarkable, indeed. What a brain fart. F. -- E-mail: info<at>simple-line<point>de
From: Virgil on 15 Dec 2006 01:45 In article <1166136192.256083.128310(a)j72g2000cwa.googlegroups.com>, "Jonathan Hoyle" <jonhoyle(a)mac.com> wrote: > Bob Kolker wrote: > > Virgil wrote: > > > > > > And it is plain that no sound mathematics can be developed unless based > > > on some axiom system as its solid foundation. > > > > Arithmetic was around long before it was axiomatized and people were > > proving theorems about integers. For example Gauss and Euler. > > > > Bob Kolker > > True, but you are ignoring Virgi's adjective "sound". Calculus existed > way back during the time of Newton and Leibniz, but you could hardly > call their use of the infinite and infinitessimals at all "sound" by > today's standards. It wasn't until Bolzano and Weierstrass made things > truly rigorous in the 19th century was Calculus anywhere near sound. > It is in fact their essential treatments that we are taught Real > Analysis today, not Newton's. (Newton's work would be barely > recognizable today with its "fluxions" and "fluents".) > > Bolzano and Weierstrass gave way to more rigor in numbers by Cantor, > and then rigor in Set Theory by Zermelo and Fraenkel. Then with the > wonderful contributions of Hilbert, Lebesgue, Godel, and others, > mathemaatics today is far more rigorous than it was over a century ago. > Even infinitessimals were consistently defined by Robinson. With the > exception of Aristotle's Logic and Euclid's Geometry, much of > mathematics would not be considered acceptable by today's standards. Actually, we today have a number of improvements on Aristoteles logic, and Euclid's axiom system had to be revamped by Hilbert to bring it up to modern standards. But they have both stood the tests of time remarkably well. > Even Guass and Euler played a bit fast and loose (although they were > considered impeccably precise in their day.) > > In ancient times, arithmetic was discovered in much the same way > physical laws were. "Hey, notice that when we do this, that always > happens..." As centuries of very hard work, mathematicians have boiled > arithmetic assumptions down to some basic axioms, and all of the > remaining theorems flow forth.
From: mueckenh on 15 Dec 2006 02:02 Dik T. Winter schrieb: > > There is a mapping. This is proven by the fact that there are more > > edges than paths. > > But the latter is simply false. Moreover, you have stated that you > had constructed a mapping. That statement was wrong? A bit sloppy. Being precise, I emphasized: I have construcuted a surjective rational relation from a subset of the set of edges onto the set of paths. This surjective rational relation proves the existence of a surjection. The notion "rational relation" is a further development of the notion "relation". While a relation connects elements of a domain to elements of a range, the rational relation connects shares of the elements of the domain to (shares of the) elements of the range. A simple example: A) Consider the relation: 1 ---> a 2 ---> b B) Now a rational relation could be: 1.1 ---> a 1.2 ---> b 2.1 ---> a 2.2 ---> b If 1.1 and 1.2 are the only shares 1, i.e., if 1.1 u 1.2 = 1 and if 2.1 and 2.2 are the only shares of 2, i.e., if 2.1 u 2.2 = 2, then one full element of the domain is available for every element of the range. Therefore B proves that a surjective mapping like A is possible, even if it cannot be found. This proof is similar to the proof of a well-ordering of the real numbers. It cannot be done (and it can even be proved that it cannot be done) but it has been proved to exist. The rational relation devised in the binary tree proves that a surjection of edges onto paths exists. In so far my assertion "there is a surjective mapping" was not quite wrong. > > But you have not proven that the tree does not contain more paths than > edges. See above. Regards, WM
From: mueckenh on 15 Dec 2006 02:07
Dik T. Winter schrieb: > > You have not understood, deplorably. The shares of every edge I use add > > to 1 edge. And every share is mapped on one path only. And every path > > gets as many shares to restate two full edges on its own. > > I have no idea of the meaning of the last sentence. > > > According to your proposal, every decimal number gets only some shares > > of the digits 1 to 9, but not as many shares as one digit is divided > > into. So not one full digit is mapped on a real number. > > Why not here? And why is that the case in your mapping? You map 10 digits on many more real numbers. Therefore not every real numbers can get the shares of a full digit. An example: If you map the digit 1 in the numbers 0.1153 and 0.3421, then you must divide the digit 1 into 2 + 1 = 3 shares, of which the number 0.1153 gets 2 shares and the number 0.3421 gets one share. But if you consider some more real numbers with, say, together having 100 digits, then it is obvious that not every real number can get a full digit (uot of these 10 digits). In the tree we have enough edges so that every path gets the shares of two full edges by 1 + 1/2 + 1/4 + ... = 2 > > > > Look at the edges. Do you deny that every edge is the beginning of that > > part of a path where it is separated from other paths? > > Right. Then you see that not more than countably many separated parts of paths can beginn in the tree. Where do you believe beginn the others which you claim to be there? Or do you claim that there are more paths than parts of paths? > > > Do you believe that paths beginn to run separated wihout any edge being > > inolved? > > No. And so what? But only looking at the paths that way you never get > the full infinite paths. If we look at last at one part of each path, then we look at all paths. If we look at all separated parts of paths, then we can be sure that we look at more objects than by looking on the whole paths. Or do you claim that there are more separated paths than sperated parts of paths? Regards, WM |