An uncountable countable set
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From: Franziska Neugebauer on 9 Jul 2006 18:45 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> > Either K is in the list (which would contradict analysis in the >> > same way as my original example), then there is an n with 10^(-n) = >> > 0. Or K is not in the list. >> >> K is not in the list. >> >> > Then there must be a position which cannot be enumerated by natural >> > numbers >> >> non sequitur. All positions are indexed by definition of decimal >> representation. > > All positions are indexed <==> K is in the list. It is a logical > equivalence for linear sets like my list. What kind of "logic" is employed here? F. N. -- xyz
From: Virgil on 9 Jul 2006 19:01 In article <1152479467.322146.132940(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > 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. > > Please note: That fact (if it was a fact) would show nothing but a > contradiction of set theory as you may obtain from he diagram above. That alleged ontradiction is only in the mind of the beholder who insists that a property of finite trees must hold for infinite trees. > > > 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. > > What happens in an any part of the infinite tree is precisely described > in the diagram. No two paths can be generated without two new edges. > Stop to run through the world of sets with your eyes wide shut. Infinite sets do not behave like finite sets. It is "mueckenh"'s eyes that are remaining "wide shut". > > > 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 > > Here it is. Every edge is mapped on that path which runs through it. That would only work if there were only one path through any edge. There are, in, fact, infinitely many paths thru each edge. Which of those infinitely many paths is to be the ->single<- image of that edge under this alleged surjective mapping? If one maps each edges on the uncountable set or paths passing through it, one gets a surjection from the set of edges to the set of uncountable sets of paths through that edge. > > Do you believe that the infinite paths exist there without consisting > of edges? Do you believe that above diagram becomes invalid somewhere? No finite diagram represents the infinite case. > This kind of collective naive blindness is required only by > set-theorists. As "mueckenh"'s attempt at a "surjection" has failed miserably, it is he who is blind.
From: Virgil on 9 Jul 2006 19:06 In article <1152479586.200077.72760(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > 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? > > 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. If those numbers represent naturals in some sort of unary notation, then 0.11... does not represent a umber at all. > > It is by far more safe to believe In mathematics, proof outweights safety. And those who only play safe discover nothing.
From: Virgil on 9 Jul 2006 19:12 In article <1152479719.404362.114580(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > 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. > > Cantor's proof allows to find the digit number n for every n. My proof > allows to find the transposition number n for every n. How can you tell what is going to happen after infinitely many transpositions which have to be taken in a particular order to assure that one knows what they are expected to do? > Cantor's > well-ordering of the rationals has to delete many rationals like 2/2 > and 6/3 or 3/6. One can surject the naturals onto the rationals without omitting any of the reducible forms of any rational. As a matter of fact, it is easier to do that that way that to biject to the reduced forms only.. In order to define the position of a certain rational > in the well-ordering, the deletion must be done by hand. This problem > becomes even more pressing when well-ordering the algebraic numbers. > All this stuff must be executed successively. And now you find that > difficult? Now you want to make us believe that in my well-determined > sequence of transpositions the ordering might be mixed up? Poor > arguing! > > > Because of the non-commutativity of permutations, your construction > > cannot use this shortcut. > > In order to give a universal law, I do not need a shortcut. But I gave > one recently: > > (1,2) > (2,3) > (1,2) > (3,4) > (2,3) > (1,2) > ... > > There is no limit process. The *only* criterion about manipulations on > countable infinite sets is whether one can determine *precisely* at > which natural number something happens: After how many steps in a > well-order of |Q the fraction 4711/235537 will appear, for instance, or > in which line of Cantor's list a certain diagonal element will be > placed and so on. And I can determine *precisely* after how many steps > the number 4711/235537 will be inserted in the order by magnitude with > all of its predecessors of the initial well-order. This is fixed and > can be calculated for any rational number. Therefore all rational > numbers are covered and will successively appear in the well-order. The > argument that there remain always infinitely many other rationals is > wrong, because by definiton the fate of each and every rational is > determined and can be calculated. It is not necessary to really carry > out any transposition. > > Regards, WM
From: Virgil on 9 Jul 2006 19:26
In article <1152479719.404362.114580(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Because of the non-commutativity of permutations, your construction > > cannot use this shortcut. > > In order to give a universal law, I do not need a shortcut. But I gave > one recently: > > (1,2) > (2,3) > (1,2) > (3,4) > (2,3) > (1,2) > ... > > There is no limit process. The *only* criterion about manipulations on > countable infinite sets is whether one can determine *precisely* at > which natural number something happens: After how many steps in a > well-order of |Q the fraction 4711/235537 will appear, for instance, or > in which line of Cantor's list a certain diagonal element will be > placed and so on. And I can determine *precisely* after how many steps > the number 4711/235537 will be inserted in the order by magnitude with > all of its predecessors of the initial well-order. Since there is no preordained position for any rational number in an arbitrary listing of the rationals, there can be no preordained number of transpositions after which it will appear in the initial ordered-by-magnitude head of the list. "Mueckenh" is wool gathering again. The most effective procedure of successive transpositions on an arbitrary infinite listing of all rationals would be a sort to bubble sort: Find the first element that is less than its predecessor and then "bubble" it towards the head of the list until it is correctly placed among what had been it predecessors. But even with this algorithm, there will always be infinitely many elements smaller that their immediate predecessor. |