Cantor Confusion
in [Math]
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From: WM on 10 May 2007 16:15 On 10 Mai, 20:09, Virgil <vir...(a)comcast.net> wrote: > In article <1178791668.851141.297...(a)y80g2000hsf.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 9 Mai, 20:47, Virgil <vir...(a)comcast.net> wrote: > > > > In fact, to establish the difference between the diagonal and any one > > > other only requires finitely many digits, which WM knows but chooses to > > > ignore. > > > This is also true for the numbers 1.000... and 0.999... > > > Why must this case be excluded from the proof? > > Because they are both excluded from appearing after the point in the > diagonal. The diagonal rule excludes both 0's and 9's after the decimal > point, so automatically cannot be equal to any number whose decimal > expansion can have them there. I ask: Why must this case be excluded from the proof? You answer: Because they are excluded. Fine. Regards, WM
From: Virgil on 10 May 2007 16:25 In article <1178827570.183346.203430(a)y80g2000hsf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 13:52, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 10, 6:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > <snip> > > > > > Give me two levels for > > > which two different paths are required. > > > > There do not exist two levels for which two different paths > > are required. > > And this is true for all levels which belong to the tree and for all > nodes any path consist of. > > > > Therefore > > > > Any finite set of paths can be replaced by a single path. > > > > Look! Over There! A pink elephant! > > > > Any infinite set of paths can be replaced by a single path. > > If you say > 1) for every path p', there is a node where p' leaves p > and > 2) there is no node where every path p' has left p > then you must be able to name a node K(n) such that for nodes K(m < > n) > at least two paths p' and p'' are required to accompany p. This is true of EVERY path and for each of its nodes in an infinite tree. For every node K(n) of P, there is one path p' which accompanies it to the next node and then branches off, and another, p'', which does not branch off until the node after that, and so on, ad infinitum.
From: Virgil on 10 May 2007 16:30 In article <1178828128.899062.206790(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 20:09, Virgil <vir...(a)comcast.net> wrote: > > In article <1178791668.851141.297...(a)y80g2000hsf.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 9 Mai, 20:47, Virgil <vir...(a)comcast.net> wrote: > > > > > > In fact, to establish the difference between the diagonal and any one > > > > other only requires finitely many digits, which WM knows but chooses to > > > > ignore. > > > > > This is also true for the numbers 1.000... and 0.999... > > > > > Why must this case be excluded from the proof? > > > > Because they are both excluded from appearing after the point in the > > diagonal. The diagonal rule excludes both 0's and 9's after the decimal > > point, so automatically cannot be equal to any number whose decimal > > expansion can have them there. > > I ask: Why must this case be excluded from the proof? > You answer: Because they are excluded. It was Cantor, or someone of his time, who created that rule, not me. The various Cantor-type rules for proper decimals all say, among other things, that the diagonal shall be made up digits other than 0 or 1. This is specifically to avoid the problem of dual representations. And it works. That WM is unhappy that it should work is irrelevant.
From: WM on 10 May 2007 16:37 On 10 Mai, 04:15, Virgil <vir...(a)comcast.net> wrote: > In article <1178739230.073383.98...(a)o5g2000hsb.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 9 Mai, 20:25, Virgil <vir...(a)comcast.net> wrote: > > > In article <1178716861.173911.234...(a)y5g2000hsa.googlegroups.com>, > > > > WM has no power to constrain mathematics to what he thinks it should be. > > > > That it is not what WM thinks it should be is by now apparent, but that > > > it is quite productive anyway is equally apparent > > > Producing what? Yes, the working mathematicians, they are quite > > productive, but they do not need transfinite set theory. Transfinite > > set theory is the most useless science ever created. > > That set theory is mathematics, and therefore not science at all, seems > to have escaped WM. In Germany mathematics belongs to "Wissenschaft" which is usually translated by science. At the university where I studied (and at most others in Germany), there is a common faculty of mathematics and natural sciences. One hundred years ago this was a world centre of mathematics and physics: Hilbert, Klein, Minkowski, Zermelo, Weyl, Born, Heisenberg, Pauli worked there. Several years earlier, also Gauss, Riemann and Dedekind. Would you call mathematics an art? Regards, WM
From: MoeBlee on 10 May 2007 16:46
On May 10, 1:15 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > I ask: Why must this case be excluded from the proof? There are different variations on the proof, depending on what interval (or even just the whole set of) reals is being proven uncountable and what base representation is being used. In any case, what the proof requires is consideration of an f that is arbitary except that is an enumeration of sequences and with not assumption that would disallow any real number (or real within the interval in question) not to be in the enumeration. Thus, by stipulating that just one of two representations of a real number may be in the enumeration, we have NOT assumed that there is a real that cannot be in the enumeration. In other word, by stipulating that we'll not allow two representations of any given real number, we're NOT barring any real number from being in the enumeration, since that real number can be in the enumeration as it has a representation that we DO allow. That you still ask about such things indicates that you do not understood the proof. Look at Suppes 'Axiomatic Set Theory' or Enderton's 'Elements Of Set Theory' for explanations that you clearly still have not grasped. MoeBlee |