Cantor Confusion
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From: Dik T. Winter on 18 Dec 2006 09:49 In article <1166452045.777425.20310(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > >> > > > All elements that can be shown to exist in the diagonal can be > > >> > > > shown to exist in one single line. [(P1)] > > > > This proposition P1 has _not_ yet been proved (shown). .... > > Upto now we have no line L_D at all since P1 has not been proved. > > > There is nothing to prove in a definition. I remember you said a > definition cannot be wrong. So it cannot be proved. P1 is not a 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: mueckenh on 18 Dec 2006 09:53 Dik T. Winter schrieb: > In article <1166361677.475397.105840(a)l12g2000cwl.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > 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"? > > > > An edge is considered to be a set of shares. You can divide it into m > > equal shares. If you divide n edges into m equal shares each, then you > > have m * n shares and you can combine m abitrary shares to have one > > full edge. As an example use euro and cent. > > And what if m is infinite? How do you divide? How do you recombine? > Suppose I have an edge that is divided in countably many shares. Now > suppose I select the even numbered shares and recombine them. Do I > now have an edge? And what when I do the same with the odd numbered > shares, do I now also have an edge? If that is not the case, why not? > And if that is the case, please explain. We need not divide anything by infinite numbers, because there are only finite numbers n in the sum 1 + 1/2 + 1/4 + ...+ 1/2^n + ... Every level of the tree has a finite number n. For finite n also 2^n is finite. > > I would not like to call such a recombination of shares an edge, because > it is *not* an edge in your original tree. I would at most call it a > collection of shares. And it is quite trivial to split an edge in an > infinite number of shares and to recombine those shares in an infinite > number of collections of an infinite number of shares. There are no infinite numbers in 1 + 1/2 + 1/4 + ... > > decimal point here. Let us denote the two equal shares of edge number 1 > > by 1#1 and 1#2. > > What is bewildering is that you are again using a theorem that would be > valid in the finite case as being also valid in the infinite case without > proof. > > > Consider you have two coins of 50 cent value each. Then you can buy > > something which costs 1 euro. > > Again, the finite case. Of course. There is only the finite case for natural numbers. There are no infinite numbers in N. > > For this sake we have mathematics, I mean really correct mathematics, > > not based on dubious infinite sets, which states that 1 + 1/2 + 1/4 + > > ... = 2. > > There are indeed texts that do state that. But what you, apparently, fail > to see is that that is not an infinite sum, but a limit in disguise. I will not discuss whatever you may call it. It is 2. That's enough. > > If AC > is true, How could it be false, if being an axiom? > there exists an Hamel basis for the reals as a vector space over > Q. And using that basis it is reasonably simple to well order the reals. Why does such an ordering definitely *not* exist? No book in any library of the world (incuding all extraterrestrical cultures) contains it, neither a formula, nor a recursion and, obviously, not a list. > > > > (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. > > > > In *mathematics*, either it can be done or it cannot be done. And if it > > has ben done, one need not forget it after abolishing AC. > > In *mathematics*, either there is a single line parallel to another line > through a point, or there is none or many. And if it has been shown, one > need not forget it after abolishing the parallel axiom. > > You fail (or are not willing) to see the similarity. There is no similarity but only the frantic attempt to keep alive a delusion. > In geometry the > parallel axiom tells us there is only single line. Abandoning that > axiom gives us different kinds of geometry. It is similar with AC. > If we accept it we get a different kind of set theory (where every set > can be well-ordered) from a version where we do not accept AC. > > Obviously with AC it has been shown that there is a Hamel basis and that > immediately leads to a well-ordering. Could you post it, please? > But when AC is false there is not > necessarily a Hamel basis, so there is not necessarily a well-ordering. > The only difference with the parallel axiom is that that axiom is a much > more direct statement. And when Lobachevsky developed his geometry he > was generally thought to be mad. That does *not* prove that, after Cohen, the advocates of AC are *not* mad. > > > > > 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. > > > > Revisit the 3rd class of the elementary school. There the teacher will > > teach you how money can be divided into smaller pieces and different > > pieces can be collected to yield a desired value, for instance euro > > 28.50 to order my book. If you found a company and every member gives 5 > > euro, then 6 members will suffice to buy the book including postage. > > He never did tell me how I can divide money among an infinite number of > persons. Perhaps you missed that lesson? > But now you are apparently stating that some of the shares can > be zero. In your tree, what shares are zero and what shares are non-zero? It is possible to divide an object into equal shares or into non-equal shares. Compare a birthday cake and some guests. In principle it is possible to give nothing to some guest. Then his share does not exist or, by modern terms, is empty. But all the shares of an edge in my tree are divided in equal parts, because of justice and because of ease of calculation. Regards, WM
From: Han de Bruijn on 18 Dec 2006 10:07 William Hughes wrote: > Han.deBruijn(a)DTO.TUDelft.NL wrote: > >>William Hughes schreef: >>> >>>Does the "potentailly infinite set" of natural numbers >>>exist? A potentially infinite set is a function on sets >>>that takes on the values true and false. The >>>potentially infinite set of natural numbers >>>takes on the value true for a set containing only >>>natural numbers, and false for any other set. >>>(i.e. is there a way of recognizing a set of >>>natural numbers?) >> >>I don't understand this question. > > Stop me at the first line you don't understand. > > [Note if you do not like the term "potentially infinite > set" substitue "recognizer".] > > A "potentially infinite set" is a function on sets. It's already here: "function" and "set". > A potentially infinite set takes on the values true and false. > > The "potentially infinite set of natural numbers", N, > is a potentially infinite set such that N(X) is true if > X is a set of natural numbers, and false if X is not > a set of natural numbers. > > Does the "potentailly infinite set of natural numbers" > exist? Han de Bruijn
From: William Hughes on 18 Dec 2006 10:07 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > >> > > > All elements that can be shown to exist in the diagonal can be > > >> > > > shown to exist in one single line. [(P1)] > > > > This proposition P1 has _not_ yet been proved (shown). > > This proposition is the definition of the diagonal. > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > contained in the diagonal. > > > > The Diagonal is unbounded thus any _assumed_ L_D is not bounded, too. > > Correct is: > If the diagonal is unbounded then any _assumed_ L_D is not bounded, > too. L_D is a line and is therefore bounded. Contradiction. Hence, the _assumed_ L_D does not exist. - William Hughes - William Hughes
From: Newberry on 18 Dec 2006 10:09
mueckenh(a)rz.fh-augsburg.de wrote: > Newberry schrieb: > > > Sorry that I joined a bit late. > Doesn't matter. > > >Are you saying that (in an infinite > > binary tree) the set of paths is uncountable but the set of edges is > > countable? > > The set of edges is obviously countable by, e.g., > > 1, 2, > 3,4,5,6, > 7,... > > As no path can separate from another one without the existence of two > more edges, the number of edges is an upper bound for the number of > paths. If it is so simple why do you still need the proof by inheritance of shares? > > Regards, WM |