Cantor Confusion
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From: mueckenh on 9 Nov 2006 05:39 Dik T. Winter schrieb: > In article <1162821418.935074.235580(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > Sorry, I misread. You insist that the limit *also* is the number of edges > > > in the infinite tree. To prove that you need transfinite induction. > > > > Then you need transfinite induction to prove that the number of > > rationals in the case of an infinite set Q is countable. > > How wrong you are. In the case of the edges you show that for finite trees > of order n: > #edges(n) = f(n) > with some function n. From that you conclude: > #edges(oo) = f(oo) > which requires transfinite induction. To prove that Q is countable you do > not need infinity at all. See the construction of a bijection I gave > from N to Q+, using simplified continued fractions. This shows that > every finite n maps to a finite q>0 and the reverse. So we have a > function g: N -> Q+ that has an inverse. And that is all that is needed > to show countability. Where does the tranfinite induction come in? There is no difference between enumerating all the rational numbers in the well-known scheme, starting at the corner of an infinite square matrix 1 2 3 6 5 4 7... and enumerating all the edges of the binary tree according to the same scheme 1 2 6 5 4 3 7 ... Or do you see a difference? But I am highly satisfied that you think it is necessary to argue the set of edges was uncountable. It shows that you understand the principle of the binary tree and its consequences. > > > > > No. The number of edges in a finite tree is without interest. We > > > > consider only the infinite tree. The set of edges there can be > > > > enumerated like the set of rational numbers, for instance. If a mapping > > > > N --> Q is defined for every element of Q, then Q is countable. The > > > > mapping N --> {edges} has been established such hat every edge knows > > > > its number. > > > > > > You are wrong. > > > > Please try to understand the concept of countability. Only in the case > > of infinite sets it is of interest. Cantor defined it without (and > > before he was knowing anything about) transfinite induction. > > Indeed. To prove countability you do not need transfinite induction. > But my remark "you are wrong" was to your statement: > "The mapping N -> {edges} has been established". > See J. H. Conway, On Numbers and Games, for a clear exposition about the > difference between the union of all finite trees and the infinite tree. Whatever Conway may say, there is no difference between enumerating all rational numbers in the well-known scheme, starting a the corner of an infinite square matrix, and the edges of the tree. In both cases you have a system with a limit which is countably infinite. Perhaps you see a difference between the union of all finite numbers n and the set N? But if that turns out necessary to keep set theory free from contradiction, then be happy with it. Regards, WM
From: Virgil on 9 Nov 2006 05:44 In article <1163067311.194466.200650(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > > If omega > n then we cannot have a diagonal with omega digits in a > > > matrix the lines of which have only n (= less than omega) digits, > > > because the digits of the diagonal are digits of the lines. > > > > That's just another reenactment of your dogmatic pronouncements. I gave > > you a proof in set theory, from axioms using only first order logic. > > You give an argument that does not use any recognizable logic and from > > no stated axioms. > > The diagonal cannot have more positions than every line. Except that if every line is finite but of length al least equal to its line number, the the diagonal /must/ have /more/positions that any line. That Eb willfully blinds himself to this necessity is just one of the many willful blindnesses that corrupt all his arguments. > see this you need not study set theory or logic or whatever. To see this you must see things that are not there and not see what is there. > It is > simply the fact which everybody accepts who has not deliberately been > blinded by assuming the existence of a whole number larger than any > natural number but counting them. WE merely assume, a la ZF, that there is a set containing all of them. We do not insist that that set is a "whole number", whatever that may be. > > This number does not exist. The set does, whether you call it a number or not. >The diagonal has not omega positions. It has "as many" positions as there are naturals, one for each natural.
From: mueckenh on 9 Nov 2006 05:49 Dik T. Winter schrieb: > In article <1162825586.068932.25310(a)m7g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > n e N was implied. If you can compare two entities in some dimension > > > > like length or weight , then they are of the same sort with respect to > > > > this dimension. > > > > > > Perhaps. But we do not talk here about length or weight. We are talking > > > mathematics. > > > > How else lengths and weight could be compared if not by mathematics? > > Weights by scales. Lengths by putting them alongside. No mathematics > involved. Comparison of numbers by size is one of the first tasks of mathematics. > But at least weight is not a mathematical entity. Length is, > but not in the way you see it. The numbers counting the units of weight or length are mathematical entities. > > > > Of course. But the usual set theory without yet specifying any models > > > > starts from this point. [everything is a set] > > > > > > Not when I followed the courses on set theory. > > > > Why didn't you? > > I did. Why do you think I did not? Because you do not know that in ZFC everything is a set, becauses you do not know that limits are inside set theory. And because you do not know why Cantor called the numbers of the first class countable by numbers of the second class. Here are some explanations of both topoi by Cantor himself. "Es ist sogar erlaubt, sich die neugeschaffene Zahl omega als ****Grenze**** zu denken, welcher die Zahlen nu zustreben, wenn darunter nichts anderes verstanden wird, als daß omega die erste ganze Zahl sein soll, welche auf alle Zahlen nu folgt, d. h. größer zu nennen ist als jede der Zahlen nu" [Cantor, Collected works, p. 195]. Die Anzahl einer wohlgeordneten Menge ist also ein Begriff der in Beziehung steht zu ihrer Anordnung; bei endlichen Mengen findet er sich offenbar als unabhängig von der Anordnung; dagegen jede unendliche Menge verschiedene Anzahlen im Allgemeinen hat, wenn man sie auf verschiedene Weise als "wohlgeordnete" Menge denkt. Die "Anzahl" lässt sich nun immer durch eine bestimmte Zahl meiner erweiterten ganzen Zahlenreihe angeben. Zum Beispiel betrachten wir die Menge aller endlichen positiven ganzen Zahlen, so ist sie in der natürlichen Folge 1, 2, 3, ..., nu... eine wohlgeordnete Menge und hat in dieser Ordnung gedacht die Anzahl: omega. Schreibt man sie aber in der Ordnung (n + 1), (n + 2), ... (n + nu), ....1, 2, 3, ..., n, so hat sie nun die Anzahl omega + n. In der Ordnung 2, 4, 6, ..., 2nu, ..., 1, 3, 5, ..., (2nu + 1), ... hat dieselbe Menge (nu) die Anzahl 2omega u.s.w. u.s.w. Jede Menge von der Mächtigkeit erster Classe ist abzählbar durch Zahlen der zweiten Zahlenklasse (II); und zwar läßt sich jede Menge von der Mächtigkeit erster Classe in solche Succession (als wohlgeordnete Menge) bringen, dass ihre Anzahl mit Bezug auf diese Succession gleich wird einer beliebig vorgeschriebenen Zahl alpha der zweiten Zahlenclasse. (Georg Cantor in a letter to Mittag-Leffler, Dec. 17,1882) Regards, WM
From: Virgil on 9 Nov 2006 05:50 In article <1163067640.060706.37510(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > The length of the diagonal is less than omega. > > > > > > > > The number of 1's in the diagonal is aleph_0. So, what you wrote is > > > > false. In fact, I have no clue what you could possibly be thinking that > > > > would lead you to make such an obviously incorrect statement. The length > > > > of the diagonal is clearly the same as the length of the first column, > > > > and you just wrote above that the length of each column is omega. > > > > > > The length of the diagonal is clearly not more than the length of any > > > line. > > > > If every line is finite but at least of length equal to is line number, > > then even more clearly, the diagonal is longer that every line. > > That is outright nonsense. How is it nonsense when that diagonal contains a position for every n to say that it is longer that any line of length less than n? It is outrageous nonsense to deny it. > Its acceptance by set theorists makes set > theory suspicious as a theory without any value. Its denial by EB makes EB a laughing stock. > > > So that WM could hardly be more wrong if he tried. > > >From the side of a set theorist I consider that as a compliment. Then EB will remain in ignorance by his own choice.
From: Virgil on 9 Nov 2006 05:53
In article <1163068762.173696.311700(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1162821418.935074.235580(a)e3g2000cwe.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > Sorry, I misread. You insist that the limit *also* is the number of > > > > edges > > > > in the infinite tree. To prove that you need transfinite induction. > > > > > > Then you need transfinite induction to prove that the number of > > > rationals in the case of an infinite set Q is countable. > > > > How wrong you are. In the case of the edges you show that for finite trees > > of order n: > > #edges(n) = f(n) > > with some function n. From that you conclude: > > #edges(oo) = f(oo) > > which requires transfinite induction. To prove that Q is countable you do > > not need infinity at all. See the construction of a bijection I gave > > from N to Q+, using simplified continued fractions. This shows that > > every finite n maps to a finite q>0 and the reverse. So we have a > > function g: N -> Q+ that has an inverse. And that is all that is needed > > to show countability. Where does the tranfinite induction come in? > > There is no difference between enumerating all the rational numbers in > the well-known scheme, starting at the corner of an infinite square > matrix > > 1 > 2 3 > 6 5 4 > 7... > > and enumerating all the edges of the binary tree according to the same > scheme > > 1 2 > 6 5 4 3 > 7 ... > > Or do you see a difference? > > But I am highly satisfied that you think it is necessary to argue the > set of edges was uncountable. It shows that you understand the > principle of the binary tree and its consequences. > > > > > > > No. The number of edges in a finite tree is without interest. We > > > > > consider only the infinite tree. The set of edges there can be > > > > > enumerated like the set of rational numbers, for instance. If a > > > > > mapping > > > > > N --> Q is defined for every element of Q, then Q is countable. > > > > > The > > > > > mapping N --> {edges} has been established such hat every edge > > > > > knows > > > > > its number. > > > > > > > > You are wrong. > > > > > > Please try to understand the concept of countability. Only in the case > > > of infinite sets it is of interest. Cantor defined it without (and > > > before he was knowing anything about) transfinite induction. > > > > Indeed. To prove countability you do not need transfinite induction. > > But my remark "you are wrong" was to your statement: > > "The mapping N -> {edges} has been established". > > See J. H. Conway, On Numbers and Games, for a clear exposition about the > > difference between the union of all finite trees and the infinite tree. > > Whatever Conway may say, there is no difference between enumerating all > rational numbers in the well-known scheme, starting a the corner of an > infinite square matrix, and the edges of the tree. In both cases you > have a system with a limit which is countably infinite. EB may finds it so, but no one else is bound by his willful blindness. |