An uncountable countable set
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From: Virgil on 10 Jul 2006 15:54 In article <1152545151.759590.283170(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > That alleged ontradiction is only in the mind of the beholder who > > insists that a property of finite trees must hold for infinite trees. > > If you are an admirer of the mysteries of the infinite, why then do you > believe that the bijection > > n <--> 2n > > or the formula > > SUM{1 to n} 1/2^n = 1 - 2^(-n) > > remains unaltered in infinity? What misleads you to believe I do? > > > Infinite sets do not behave like finite sets. It is "mueckenh"'s eyes > > that are remaining "wide shut". > > > That would only work if there were only one path through any edge. > > > > There are, in, fact, infinitely many paths thru each edge. > > The first edge is mapped on the first path. Which of the infinitely many paths through that edge is the "first" one? > If this splits in two, 1/2 of the first edge in addition to the new > edge is mapped on the new paths. You are not allowed to split edges. Besides which, the first edge has already been entirely used up. If the new paths split, 1/4 of the first edge and 1/2 of the > second and all of the third are mapped on each path. Do you know what a > geometric series is? What do you object to this mapping? If one can split edges into infinitely many peices one can equally split paths. it is only unsplit edges and unsplit paths that are to be matched up in injections, surjection or bijections.. > > > > Do you believe that the infinite paths exist there without consisting > > > of edges? Do you believe that above diagram becomes invalid somewhere? AS soon as you split edges, your analysis crashes. > > > > No finite diagram represents the infinite case. > > But a finite brain can think of the infinite. If it is not as devoid of mental capacity as "mueckenh"'s, yes!
From: Virgil on 10 Jul 2006 16:02 In article <1152545261.586900.23860(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > Precisely where does the Cantor diagonal proof fail? > > > > > > In this and infinitely many infinite lists like this: > > > 0.0 > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > > > > In an infinitely long list the diagonal number 0.111... is not > > > distingiushed from all the list numbers. > > > > If those numerals represent fractions then the thing represented by > > 0.111... is also a fraction, but is not in the list. > > All positions which can be enumerated are in the list, by definition. > Hence, the decimal representation of 1/9 does not exist. Then how is it that so many people use it? And all sorts of other non-existent decimal representations? Each of these allegedly nonexistent representations is also a representation of a convergent geometric series, and is (by custom if not more explicitly) taken to represent the limit value of that convergent series. That is no more arbitrary than representing natural numbers in decimal notation.
From: Virgil on 10 Jul 2006 16:08 In article <1152545532.650765.321300(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > But in order to prove that the diagonal differs from every entry, there > > > a limit is required but not available. > > > > Not by me. I can easily do it without any limit process at all. > > Because you are a blind believer. Because I can prove that there is a non-zero difference between the the diagonal and any real in the list (for the number in positions n, the difference is >= 1/10^n > 0), for all n simultaneously. > > > > > The missing limit has not been remedied but has only been put aside. My > > > objection remains: But that there is no satisfactory limit > > > consideration becomes clear from the following: We know that 0.999... = > > > 1.000... This leads to the result that a change of 1 in the limit where > > > the digit number goes to oo does not have the effect which would be > > > required in order to distinguish the diagonal number from the list > > > numbers. > > > > Cantor's rules avoid that particular pitfall by never using either 0 or > > 9 in diagonal constructions. > > Cantor did *not* consider this pitfall. But you would have believed him > nevertheless. Whether Cantor himself did it or someone else, is irrelevant. The pitfall is avoidable simply by never using a 0 or 9 in the construction.
From: Dik T. Winter on 10 Jul 2006 19:14 In article <virgil-E2E7E4.14081410072006(a)news.usenetmonster.com> Virgil <virgil(a)comcast.net> writes: > In article <1152545532.650765.321300(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: [about dual representations]: > > > Cantor's rules avoid that particular pitfall by never using either 0 or > > > 9 in diagonal constructions. > > > > Cantor did *not* consider this pitfall. But you would have believed him > > nevertheless. > > Whether Cantor himself did it or someone else, is irrelevant. > > The pitfall is avoidable simply by never using a 0 or 9 in the > construction. In his first diagonal proof Cantor did not consider that pitfall, because in that proof it *is* not a pitfall. His first diagonal proof starts with ordered lists of elements 'm' and 'w', so E_k = (..., E_k[i], ...) where E_k[i] is m or w. He shows that the set of such things is not countable by constructing a new element E_0, given a list E_1, E_2, ..., where E_0[k] = if E_k[k] is m then w else m. It is only when you look at them as binary numbers (as Mueckenheim does) the pitfall comes in. But in his proof Cantor does *not* look at them as binary numbers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 10 Jul 2006 23:31
In article <1152384241.144334.238730(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. It will be necessary if "mueckenh" claims a unary representataion of w. > > > 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. How does it do that when 0.111... has to be two numbers at once? > > Regards, WM |