Cantor Confusion
in [Math]
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From: WM on 2 May 2007 07:54 On 1 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: Dear Dik, you are invited: http://groups.google.nl/group/AOTI/browse_thread/thread/da0a93f42305b2c8 Gruß, WM
From: Virgil on 2 May 2007 14:15 In article <1178106864.347366.36570(a)c35g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > http://groups.google.nl/group/AOTI/browse_thread/thread/da0a93f42305b2c8 WM's "intercession" of two sets is merely having each set of two sets dense in their suitably ordered union, and does not eliminate the very real differences between sets incapable of being bijected.
From: WM on 3 May 2007 06:43 On 30 Apr., 19:28, Virgil <vir...(a)comcast.net> wrote: > In article <1177936379.128093.174...(a)n76g2000hsh.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > Anyhow. The number of path bunches surpassing the number of nodes is > > zero. If any infinite path p exists, then the bunch {p} does exist > > too, and {p} is a path bunch. If, on the other hand, {p} does not > > exist, then it is ridiculous to try to count the number of paths. > > So that WM is essentially saying that there is no such thing as an > infinite binary tree. No meaningful calculating with infinite numbers is possible without such contradictions arising. > > The problem is that he has no axiom system in which he can prove his > conclusion whereas we have several in which we can prove that such trees > do exist and have uncountably many paths in them. > > The ZF system and the NBG system, for example, both not only allow, they > require that CIBTs exist and that the set of paths in them are > uncountable. That is their problem: These systems are self-contradictory. > > What axiom system does WM claim allows him to prove otherwise? 2-1-1+2-1-1+2-1-1+-... is never much more than 2. Regards, WM
From: WM on 3 May 2007 06:46 On 30 Apr., 19:33, Virgil <vir...(a)comcast.net> wrote: > In article <1177936511.069913.212...(a)e65g2000hsc.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > On 27 Apr., 22:06, Virgil <vir...(a)comcast.net> wrote: > > > > > If something is valid for all elements, then it does not change if the > > > > number of elements is infinite. > > > > What is true of elements one at a time need not be true for infinite > > > sets of them. > > I.e., properties of members of a set need not be inherited by the sets > themselves. > > > Why then do you believe that Cantor's diagonal proof is true? > > Cantor's proof does not say anything is true for a set itself, only for > each member of a set, so it is not the same thing. Cantor's proof says that all members of the set R are more than all members of the set N. This is a wrong conclusion. Only for each diagonal number one can say, at most, that it was not in the list while it, nevertheless belongs to a countable set. Regards, WM
From: WM on 3 May 2007 06:55
On 30 Apr., 20:04, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Apr 30, 5:35 am, mueck...(a)rz.fh-augsburg.de wrote: > > > On 27 Apr., 22:06, Virgil <vir...(a)comcast.net> wrote: > > Why then do you believe that Cantor's diagonal proof is true? > > You'd have to define 'true proof'. Meanwhile, with just some basic > knowledge of predicate calculus and set theory, one can see that > Cantor's diagonal argument is formalizable into a proof in first order > logic from the axioms of Z set theory. There is an implicit assumption entering which is wrong. The difference of 1 between the digits of two numbers is insufficient to distinguish these numbers in the limit n --> oo. lim(n-->oo) 10^-n = 0. Only this limit makes power series of digits converge and makes real numbers exist. But Cantor's proof forgets that. > If you understand the basic > material, there is not much to believing, since it is a simple of > matter of observing that a certain sequence of formulas that satisfies > the definition of 'first order proof from the Z axioms' does exist. > And as easily one can see that the difference between number of paths and number of nodes in the binary tree is given by 2-1-1+2-1-1+2-1-1+-... which is not much more than 2. If you understand the basic material, there is not much to believing. Regards, WM |