Cantor Confusion
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From: Virgil on 15 Dec 2006 17:46 In article <1166185295.562736.84050(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > > No I simply state that the diagonal proof fails in case of the tree. > > > > AS the diagonal proof is not allied to trees, so what? > > > Its contradiction is. Show me! > > > > > > We can simply count the edges. They are countable. So we an count the > > > biginnings of separated parts of paths (because every edge is a > > > beginnng of the separated part of a path, notwithstanding which it may > > > be). > > > > > > Every finite beginning of a set of paths, however long, is the beginning > > of as many paths as the set of all paths. > > > > So there are as many paths following any such beginning as there are > > paths altogether. > > And there are as many beginnings following such a beginning. if a beginning is either a node or an edge, there are fewer beginnings that unending paths from such a beginning in an infinite binary tree. > > > > > Therefore the beginnings of separated parts of the paths are > > > countable. > > > > > > It is the endings that are not countable. > > It is the endings, which do not exist. Each path has a head node and a tail consisting of everything else. The number of heads in an infinite binary tree is countable, the number of endless tails from any node is equal to the number of endless binary strings which is not countable. > > > > > Do you deny that every edge is the beginning of that > > > part of a path where it is separated from other paths? > > > > I deny that it has any relevance to the "number" of paths. > > To state it clearly: You have to believe (or at least to assert to > believe) that there are more separated paths than beginnings of > separated paths in the tree. I have no idea how a separated path is different from a path. Every path in a tree has the same initial root node, so that there is only one beginning common to all paths. If WM is referring to endless sub-paths whose initial node is not the root node, the set of all such having the same initial node bijects easily with the set of all full paths in the original tree. > > Without some assumptions (other than the rules of logic) one > > cannot deduce anything about anything. > > So one must start with SOME assumptions. > > No. One can start from some observations like 1+1 = 2. That assumes meanings for "1" and "+" and "=" and "2".
From: Jonathan Hoyle on 15 Dec 2006 17:56 > In other words, Jonathan, Aristotelian syllogistic inference only > yields truisms not truth. It can't tell us what the truth of anything > actually is only whether or not it's true in relation to constituent > premises of equally problematic truth. I cannot imagine how anyone would have expected it any different. OF COURSE the truth of a statement is based upon the truth of its premises. I don't believe Aristotle thought his work failed to "produce the truth he hoped it would yield". I think you have unrealistic expectations from logic. Were you imagining that Aristotle was planning to come up with a system which would determine the truth or falsity of a statement A, irrespective of any premises? What if A = "God exists."? Or how about A = "John is handsome."? > The truth of conclusions is not > inherited from the truth of constituent premises as syllogistic > inference implies and Aristotle thought. The truth of any conclusion > must be demonstrated in its own right by demonstrating the falsity of > alternatives by finite tautological regression to self contradictory > alternatives. Ridiculous. How would you apply such a principle to the Reimann Hypothesis? Or to something non-mathematical, like a non-reproducible historical fact like "the U.S. went to the moon." Without *some* assumptions (or in mathematics, axioms), you cannot arrive to any conclusions of truth. (In the aforementioned example, there are conspiratorialists who deny some of the more basic assumptions that you might not realize you are even making.) Perhaps I misunderstand where you are coming from, but I very much doubt that Aristotle held the expectations you had for logic. Jonathan Hoyle Eastman Kodak
From: Dik T. Winter on 15 Dec 2006 22:22 In article <1166166464.741143.287220(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. And again you answer not to all my question but only to those you can answer. What is the answer to this? > > > 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. Makes no sense. You map edges to many more real numbers. Therefore not every real number can get the share of a full edge. What does that *mean*? > An example: I would prefer it when you provide definitions rather than examples. > 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). Eh? Not obvious at all. Moreso because the terms you are using are not defined. > 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 What do you *mean* with the term "the shares of two full edges". As stated it is plain nonsense. > > > 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? The last, and that is easily proven, when you think about parts of paths as initial finite segments of paths. And indeed, there are countably many initial finite segments of paths in the tree, as I have shown already. > > > 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? Again, lacking any definition of the terms you are using. But it all boils down to your denying the axiom of infinity. You state that ZF is inconsistent but in all your proof you are using the negative of the axiom of infinity. It should be pretty obvious that ZF plus the negation of the axiom of infinity is inconsistent. That does *not* show that ZF is inconsistent. But apparently you are not able to see that you use the negation of that axiom in your reasonings. -- 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 15 Dec 2006 22:46 In article <1166166146.392714.316100(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Getting weirder and weirder. What is "a surjective rational relation"? > 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: Again, what are "shares"? > 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. Makes barely sense. What is 1.1 united 1.2? I have no definition, so I have no idea what that means. And even going into your reasoning, I see still nothing that allows the "then one full element of the domain is available for every element of the range". I would like to ask for a proof. You are still reasoning in the finite case. Whether it is applicable in the infinite case remains a question. > 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. You are entirely wrong here. (1) Axiom of choice valid: it can be done (and I think it has been done). (2) Axiom of choice not valid: you can prove that it can not be done. But you confluate the two together and apparently do not see the difference. > 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. Not so. Because you are using a lot of terms that lack definition. -- 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 15 Dec 2006 23:11
In article <JABMHq.8D3(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <1166167313.573914.50760(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > cbrown(a)cbrownsystems.com schrieb: > ... > > > 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". > > > > Unfortunately this link was truncated. I try to post it again. > > I do not think that link makes much sense for somebody who does not > understand the German sentence above. However, I do understand why > you post that link. How much did publication cost you? My estimate > is about EUR 1000. For those who wonder. Shaker Verlag is nothing more than a printing company. It was started to aid universities to publish their books in a nice format and to publish thesises. In addition to the printing it does something about ISBN's, some publishing and so on. But they appear to be going into the field of vanity press. Whatever you wish to publish, they will do it, if you pay what they ask you. If you have a manuscript of about 150 pages, they will publish with a print run of 200, for about EUR 1000. No review, no editing. Just printing. Perhaps a good idea for JSH (they publish in English, and I think in whatever language you can imagine). Oh, and as a nicety, the author will receive parts of the sale cost for each book outside the initial print run. It is 10%, but I do not know whether that is before taxes or after. So my estimated cost for WM would be something like EUR 1000. The book is 168 pages and costs EUR 28.50, so WM needs about 350 regular sales to recoup his investment. See <http://www.shaker.nl/english/Info/Publicatieconcept.asp> for a statement from the Dutch branch. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |