Cantor Confusion
in [Math]
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From: WM on 11 May 2007 04:57 On 10 Mai, 23:03, William Hughes <wpihug...(a)hotmail.com> wrote: > Note: you can enumerate an infinite set of nodes. Let us assume that. > > For every node n belonging to the set of nodes that you can enumerate, > one path p'(n) is sufficient to accompany p up to that node. > p'(n) is not the same for every node. For which node is another path required? Regards, WM
From: WM on 11 May 2007 04:59 On 11 Mai, 00:07, Virgil <vir...(a)comcast.net> wrote: > > 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? > > Yes! And what about theology? Is it a science or an art or something else? Regards, WM
From: WM on 11 May 2007 05:07 On 11 Mai, 00:22, Virgil <vir...(a)comcast.net> wrote: > In article <1178830047.112233.306...(a)p77g2000hsh.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 10 Mai, 21:36, Virgil <vir...(a)comcast.net> wrote: > > > In article <1178806498.433405.226...(a)o5g2000hsb.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 10 Mai, 03:59, Virgil <vir...(a)comcast.net> wrote: > > > It is irrelevant what it is according to your opinion or "in the first > > place". > > English definitions of function, at least those in the more rigorous of > texts, insist that they be sets of ordered pairs, and "single-valued" in > the sense that if <a,b> and <a,c> are members then b = c, but do not > insist on there being any rule by which they are defined. If a set of ordered pairs is defined, there must be a formula or whatever you may call it, defining it. Otherwise it is undefined. Further the first elements and the second elements form sets. Without such sets there are no triples aRb etc. > > Once defined, one can then define other related sets, such as a domain, > codomain or range, but these are only consequences of the original > definition, as are such properties as being injective or surjective. > > I said that a function consists of formula, domain and range. > > There need not be a 'formula', in any formal sense, in evidence in order > to have a function, but one does need a set of ordered pairs. And just this set is defined by the formula. > > > This is true and provable > > It is only provable when it is assumed. It is provable that domain and range exist, if you have a binary relation. Nothing else has to be assumed. See Exercise 2.1 of H&J: Let R be a binary relation; show that dom R c U(UR), ran R c U(UR). Conclude from this that dom R and ran R exist. Regards, WM
From: WM on 11 May 2007 05:35 On 11 Mai, 00:25, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 10, 6:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 9 Mai, 22:43, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > >...there must be such a path. > > > >Why not call it p''. > > > > because the name p'' has already been used > > > for a path that does branch off from p. > > > No, it has not. p'' is, by definition, that path which is with p also > > at the next node, wherever you rest. > > Funny, I thought it was different. No. That was my original definition which you changed. But if you claim to want to use p'' only according to your definition, then I'll take p* as: Definition: p* is a path in the binary tree which is with path p at the next node K(p, n+1) in the binary tree, whatever node K(p, n) you investigate. Lemma: In the binary tree for every path p there exists at least one path p* as defined. Proof: Choose a node K(p, n) of path p in the binary tree. There is a path p* which is with p at node K(p, n+1) too such that K(p, n+1) = K(p*, n+1). This path p* has been with p at all nodes m < n too. The choice of n is arbitrary. Therefore the Lemma holds for all nodes K(p, n) of p with n in N. Remark: This proof is independent of the number of nodes. Regards, WM
From: WM on 11 May 2007 06:22
On 10 Mai, 22:30, 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. No, it was not Cantor. It was someone who recognized that lim[n-->oo] 10^-1 = 0 in case of dual representation, i.e., 1.000... = 0.999... Unfortunately he did not recognize that he same holds for the difference between the diagonal number and a list entry in case of this limit. The reason is clear. Nobody ever considered the case that a real number is changed in the limit lim[n-->oo] except the natural "dual representation". And Cantor's proof deceives the reader by stating that every digit has a finite index. In fact, that is true, but it is also true for dual representation. There it is not sufficient to yield one and the same number. > > 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. > Why is there a problem if only digits at finite places are changed? Why is there a problem with lim[n-->oo] 10^-1 = 0 in one case but not in another one? Regards, WM |