An uncountable countable set
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From: mueckenh on 14 Jul 2006 09:19 Virgil schrieb: > > This is a set and it exists as well as any other infinite set. Not more > > and not less. > > What set? All the transpositions required form an infinite set. As such it should be possible to be exhausted. Regards, WM
From: mueckenh on 14 Jul 2006 09:22 Virgil schrieb: > > > For infinite binary trees each such list of branchings is an infinite > > > sequence. And there are uncountably many possible such infinite > > > sequences. > > > > But we know that there are nearly twice as much edges. > > We only know that for finite trees. "Twice as much" doesn't work for > infinite sets. But less than twice also does not work. > > > There are > > infinitely many of them too. And the the set of edges is countable. > > Then show me an ennumeration of them. I wait until you will have learned to calculate with fractions. Then you can see that every path carries two edges. Regards, WM
From: mueckenh on 14 Jul 2006 09:24 Virgil schrieb: > > 0.111... cannot be indexed by the members of any natural. Does 0.111... > > have more 1's than can be indexed by any natural? Or what prevents it > > from being indexed by any of the naturals? > > What cannot be indexed by any member of N is N itself, but N itself can > be "indexed" by the members of Succ(N) = N \union {N}. Doesn't N consists of naturals only? Can't each natural index itself? What remains, if each natural has indexed itself? > > Each ordinal has a successor, even limit ordinals like N, and any > ordinal can be indexed by the members of its successor. Each ordinal can be indexed by itself. 1 is indexed by 1. 1,2 is indexed by 1 and 2. 1,2,3 is indexed by 1,2 and 3 and so on. But all naturals are not indexed by all naturals? Extremely ridiculous! Extremely! Regards, WM
From: mueckenh on 14 Jul 2006 09:29 Dik T. Winter schrieb: > > This definiion is > > wrong. I show this by the fact that 0.111... is not in the set of unary > > reprsetations of all natural numbers. aleph_0, the number of 1's in > > 0.111... is *not* the cardinal number of |N. > > Proof, please. Define 0 *+ 0 = 0, 0 *+ 1 = 1 *+ 0 = 1 *+ 1 = 1. Do the following addition *+ 0.1 0.11 0.111 .... What is the result? 0.111... This number is not among the numbers to be added. But it must be, if it is different from any list number and if the result of the addition is 0.111... . Its absence does not matter as its presence does not matter. Add by *+ 0,1 0.11 0.111 .... 0.111... The result is the same. Hence, if 0.111... exists, it is in he list and is not in the list. What kind of objects have the property to be and not to be? > > > K cannot have infinitely many digits if all can be indexed by finite > > numbers. Infinite is larger than any finite number. Therefore every > > list number taken will not be capably of indexing K. > > 0.111... - 0.111...1 = 0.000...0111... > > You mean covering here. And you are right, K is not in the list. I have > never argued that it is. Still, what is the problem? What cannot be covered by naturals cannot be indexed by naturals. > > What do you mean with "cannot be indexed"? A position to be enumerated by a natural ordinal number (starting from the first ,1) > > You use the word "indexed" in a strange way. You mean "covered". I never > assume that 0.111... can be "covered" by some An. Nevertheless, *each > digit position can be indexed by some An". As 0.111... consists of nothing else than 1's at each position, covering of each position means covering of the whole number. Hence your standpoint involves a contradiction. Regards, WM
From: mueckenh on 14 Jul 2006 09:35
Dik T. Winter schrieb: > > These words are uninteresting. Transpositions or exchanges of two > > elements are clearly defined. Infinitely many are admitted by Cantor. > > This is wrong. Not any sequence of transpositions is permitted. What *is* > permitted is *transformations that can be written as a sequence of > transpositions*. Because if you admit *any* sequence of transpositions > you may destroy the order-type. Works, p. 214: Die Frage, durch welche Umformungen einer wohlgeordneten Menge ihre Anzahl geändert wird, durch welche nicht, läßt sich einfach so beantworten, daß diejenigen und nur diejenigen Umformungen die Anzahl ungeändert lassen, welche sich zurückführen lassen auf eine endliche oder unendliche Menge von Transpositionen, d. h. von Vertauschungen je zweier Elemente. No word about sequence or permutation at this place. > > > The reordering of the naturals (0, 1, 2, ...) to (1, 2, 3, ..., 0) can > > > be written as an infinite sequence of transpositions, but it changes the > > > ordinal number. But is it a permutation? > > > > Unimportant for Cantor's statement. > > It *is* important. Because if that falls under Cantor's definition of > "transformation" and "permutation", Cantor's statement is trivially false. Of course it is false, whether there is a sequence or not. My transpositions form a sequence. One cannot do better. > Eh? You assume that Cantor's conclusion is true for your sequence of > transpositions. I state that Cantor's conclusion is false for your > sequence of statements. Cantor's conclusion is correct, with no doubt, if infinite sets like the set of my transpositions do exist.> > > > I know your interpretation is wrong. > > What part of my interpretation is wrong? The implication that Cantor's statement could be correct (if special constraints were obeyed by permutations) with the implication that infinity could exist. > > > But that does not at all affect my proof, because, as Virgil, > > emphasized frequently, there is no transposition which changes the > > well-order to normal order while maintaining the initial enumeration (= > > well-order). Would infinitely many transpositions be possible, this > > would necessarily happen. > > No. There would still not be a single transposition at which well-order > changes to non well-order. As in lim{n -> oo} 1/n, there is not a > single natural at which 1/n becomes 0. Therefore it never becomes 0 but always remains > 0 like 0.11... never becomes 1/9. Regards, WM |