An uncountable countable set
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From: mueckenh on 8 Jul 2006 14:44 Dik T. Winter schrieb: > > > > > > Either the diagonal number 0.111... is not distinguished from all > > > > > > finitely large numbers of the list > > > > > > 0. > > > > > > 0.1 > > > > > > 0.11 > > > > > > 0.111 > > > > > > ... > > > > > > then Cantor's proof fails. > > > > > > If we assume that that list contains natural numbers in some unary notation > > > (as I think you do) than: > > > > > > > > > Or 0.111... is distinguished from all finitely large numbers of the > > > > > > 0.111... is not a natural number in unary notation. So it is inherently > > > different from all elements of the list. > > > > OK. it is the unary representaton of omega. > > Interesting. What is the unary representation of w+1? As we will see that the unary representation of w does not exist, it will not be necessary to look for the unary representation of w + 1. > But more clear, you > admit that it is not a natural number, and so inherently different from all > elements of the list. This would be necessary, yes. > > > > > "To be different" means for all unary representations of n > > > > An : 0.111... - n =/= 0 > > > > > > How do you propose to define that subtraction when 0.111... is not a > > > natural number? > > > > I chose just this number because it is also the decimal representation > > of 1/9. > > Interesting, although it makes no sense. It allows us to subtract: 0.111... - 0.111...1. Regards, WM
From: Virgil on 8 Jul 2006 14:50 In article <1152382404.429097.306020(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > | > > > o > > > / \ > > > o o > > > / \ / \ > > > > > > > When every path contains more than any finite number of edges, then > > there are "more" paths than edges. > > Then there are many edges, of course. But why should there be many more > paths? Because one can biject the set of edges with the naturals and biject the set of paths with the power set of the naturals, and there are "more" members in any power set that in the set itself. > > That is even harder logic. > > It is nonsense to believe that the above diagram would ever become > invalid. But the diagram shows that there is no branching without two > new edges. Hence there *cannot* be more paths than edges. It is simply > impossible as long as paths consist of edges. It is only possible to have more edges than paths when one cannot biject the set of paths with the power set of the set of edges. No finite tree represents what happens in an infinite tree. > That is hardest logic, so > hard that even HE HIMSELF could not neglect it. But Virgil himself > believes to succeed? A I have, myself, analysed the indirect bijection between the set of paths in the infinite binary tree and the power set of the set of paths, and found it sound, I reject anyone else's unproven claims to the contrary. If "mueckenh" wishes to claim more edges than paths, or even as many edges as paths for such ->infinite<- trees, he must either construct an explicit injection from the set of paths to the set of edges or a surjection in the reverse direction > > > > I , and others, have previously demonstrated a bijection between the set > > of all infinite paths in a complete binary tree and the set of all > > subsets of N, P(N). > > > That would merely show that set theory is self-contradictory. No! It would only show that "mueckenh"'s beliefs about that theory contradict the facts of that theory. > > So that "mueckenh"'s claim that there are "more" edges than paths, is a > > claim that there is a surjection from N to P(N) (or equivalently a an > > injection from P(N) to N) but no injection from N to P(N). > > > If I said "more" this was in the usual sense. According to set theory > it should read "not less edges than paths". Then "mueckenh" is claiming a surjection from edges to paths. While this is quite possible for finite binary trees in which there is a finite maximal length to each path, and each path has a ->last<- edge, it is not for infinite binary trees in which each path contains infinitely many edges and no path contains any last edge. "Mueckenh" is conflating properties that only hold in finite cases with what will hold in the infinite case.
From: mueckenh on 8 Jul 2006 14:53 David Hartley schrieb: > In message <vmhjr2-E08407.11521207072006(a)news.usenetmonster.com>, Virgil > <vmhjr2(a)comcast.net> writes > >In article <1152282125.757193.28320(a)m73g2000cwd.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > >> There is no limit other than in Canrtor's diagonal. EVERY set initially > >> indexed by natural numbers will remain indexed by natural numbers. And > >> it will unavoidably become ordered by size too. > > > >At best you have proved that a countable subset of the rationals have > >ben restored to their usual ordering, but you have nowhere shown that > >your process ever re-orders ALL of them. Cantor has nowhere shown that his definition covers all natural numbers. > > > >After each of your countably many transpositions, there is still a > >countably infinite subset of the rationals left which are not in natural > >order. That is a necesssary consequence of the entire set being > >originally well ordered. But in the moment of the definiton, the result is determined for all rationals as the well-order is determined for all rationals, though in fact only a small set can be written. It is the principle which counts. Therefore your arguing is invalid. > > Suppose x, y are any two rationals. Suppose they are indexed by m and n > in the first enumeration, and let k = max(m,n). From the k-th stage on, > x and y are in their natural order. Under any reasonable definition of > the limit of a sequence of orderings, here it will be the natural > ordering. And it will be covering all rationals, as soon as the transposition principle is given. > > (WM's fallacy is his claim that the (order-preserving) indexation by the > naturals carries over to the limit.) Cantor's fallacy is that the indexation by the naturals carries over to the limit. Regards, WM
From: Virgil on 8 Jul 2006 15:25 In article <1152383284.249406.311540(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152191557.066779.80730(a)p79g2000cwp.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > In article <1152133409.564175.47720(a)p79g2000cwp.googlegroups.com>, > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > > > > 0.1 > > > > > > > > > 0.11 > > > > > > > > > 0.111 > > > > > > > > > ... > > > > > > > > > 0.111...1 > > > > > > > > > ... > > > > > > > > > > > > > > > > > > But in this list the number 0.111... is not contained. Hence > > > > > > > > > not > > > > > > > > > all > > > > > > > > > of > > > > > > > > > its digits can be identified. > > > > > > > > > > > > > > > > You have just proved that your example supports Cantor's > > > > > > > > theorem > > > > > > > > by > > > > > > > > producing a number, 0.111..., not in your own list. > > > > > > > > > > > > > > All digits of a number must be indexed by natural numbers. > > > > > > > Otherwise > > > > > > > they cannot be identified. All digits which can be identified are > > > > > > > pesent in the list which contains all unary representations of > > > > > > > naturlal > > > > > > > numbers. > > > > > > > > Those are not natural numbers at all in the list, but decimal, or > > > > other > > > > base, fractions. > > > > > > These are unary representations of natural numbers: 1 = 0.1, 2 = 0.11, > > > 3 = 0.111, ... > > > > > > What is your point in having only a sequence of natural numbers? > > > > No one has claimed that the set of naturals is uncountable and no one > > has claimed that a sequence of distinct natural numbers converges to a > > natural number limit, so that a Cantor "anti-diagonal", or a limit value > > for a sequence of distinct naturals is irrelevant. > > If Cantor's proof fails once, it fails once and for all, because it is > an impossibility proof. Due to the claim it is impossible to have the > diagonal in the list. Precisely where does the Cantor diagonal proof fail? Given any list of reals, the Cantor rule provided a method of finding a real number appearing nowhere that list. A slight modification allows one to construct a separate and distinct real nonmember of a given list of reals for every member of P(N). The only failure I see here is "mueckenh"'s failure to understand what the Cantor theorem and its 2nd proof say.
From: Virgil on 8 Jul 2006 15:43
In article <1152383363.139355.165360(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Therefore we have a well-ordered set simultaneously ordered by size, > "at the end" of finity. When you get to the "end" of that which has no end, send us a postcard. Cantor's proof avoids this need to "reach the end" by giving a general rule which works independently for each n in N without reference to any other member of N. This allow the simultaneous creation of all the digits. Because of the non-commutativity of permutations, your construction cannot use this shortcut. |