Cantor Confusion
in [Math]
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From: William Hughes on 7 May 2007 07:33 On May 7, 7:24 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 7 Mai, 00:59, William Hughes <wpihug...(a)hotmail.com> wrote: > > > Look! Over there! A pink elephant! > > That seems necessary to make us believe that a paths that shares > every node with another path does not share every node with another > path. But it is not sufficient. > > Regards, WM Crank rule #1: When losing an argument, change the subject. Every node, n, of path p is shared with another path p'(n). Note that p'(n) can be different for every n. Look! Over there! A pink elephant! Every node, n, of path p is shared with another path p'. p' never changes. - William Hughes
From: WM on 7 May 2007 07:45 On 7 Mai, 13:33, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 7, 7:24 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 7 Mai, 00:59, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > Look! Over there! A pink elephant! > > > That seems necessary to make us believe that a paths that shares > > every node with another path does not share every node with another > > path. But it is not sufficient. > > > Regards, WM > > Crank rule #1: When losing an argument, change the subject. You are very good at that topic. > > Every node, n, of path p is shared with another path > p'(n). Note that p'(n) can be different for every n. Note that p'(n) need not diffrent for every n. > > Look! Over there! A pink elephant! > > Every node, n, of path p is shared with another path > p'. p' never changes. > For every n there is a p' which has not changed from 1 to n. Regards, WM
From: William Hughes on 7 May 2007 08:10 On May 7, 7:45 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 7 Mai, 13:33, William Hughes <wpihug...(a)hotmail.com> wrote: > > > On May 7, 7:24 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 7 Mai, 00:59, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > Look! Over there! A pink elephant! > > > > That seems necessary to make us believe that a paths that shares > > > every node with another path does not share every node with another > > > path. But it is not sufficient. > > > > Regards, WM > > > Crank rule #1: When losing an argument, change the subject. > > You are very good at that topic. > > > > > Every node, n, of path p is shared with another path > > p'(n). Note that p'(n) can be different for every n. > > Note that p'(n) need not diffrent for every n. However, it cannot be the same for all n. > > > Look! Over there! A pink elephant! > > > Every node, n, of path p is shared with another path > > p'. p' never changes. > > For every n there is a p' which has not changed from 1 to n. Call it p'(n) Look! Over there! A pink elephant! p'(n) never has to change. - William Hughes
From: Virgil on 7 May 2007 11:46 In article <1178532990.891073.253450(a)y5g2000hsa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 6 Mai, 20:16, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 6, 10:53 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > Fine. That means, the set is countable but there is no bijection with > > > N definable. > > > > No it means that while a bijection exists it is not finitely > > definable. > > The bijection with this countable set is not finitely definable. > Why then do you require a finitely definable bijection between paths > and nodes in the tree? > > Regards, WM You are the one pushing "finitely definable", whatever that means. But as the construction of the Cantor diagonal is finitely defineable, and finitely defined, there should be no objection to it.
From: Virgil on 7 May 2007 12:06
In article <1178533278.892846.274640(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 6 Mai, 22:10, 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. > > > > > > What WM is describing is not a difference between two fixed numbers but > > > > between two sequences of numbers whose differences converge to 0, the > > > > limiting difference is indeed zero but that is totally irrelevant, > > > > > not for the sequence of digts of the diagonal number and any other > > > entry of the list. > > > > Since the "diagonal" differs from each listed member at some finite > > place value, what happens at later place values is irrelevant to whether > > they are equal or different. > > The number of separated paths is les than the number of nodes at each > finite level. What happen after each finite level is irrelevant. The number of paths, or more properly uncountable sets of paths, distinguishable at any level is, like the level itself, finite, but there are infinitely many levels, and what happens at any one level is not the end. In order to consider the whole tree one must consider all infinitely many levels. For every complete finite tree (all paths of equal length), the number of paths equals the number of subsets of the set of non-leaf levels. Each path being identified with the set of levels from which it branches left. WM argues elsewhere that what holds in the finite cases must carry over to the infinite case. SO why does WM think that this suddenly becomes false in the limit as the number of levels becomes infinite? WM seems to have one view of limits when he thinks they will support his opinions, and the opposite view when they oppose them. > > > > While there may be some subsequence of the original which converges to > > the diagonal, no member of that subsequence EQUALS that diagonal, which > > is all that is required. > > That is not different in case of 0.999... and 1.000... > Why must it be excluded? Who says it is? Any decimal expansion which has no 0 or 9 in it must be different from any decimal expansion having any of either, and the so called Cantor diagonals for decimal expansions have no 0's or 9's. > > > > > > > > > > In base b, numbers of form m/b^n have two expansions. But excluding > > > > these, a difference of 1 or more in any digit position differentiates > > > > between two different reals. > > > > > It does only differentiate the reals at a finite index. But at a > > > finite index it differentiates as well between 0.999... and 1.000.... > > > So, why do you want to exclude these but not those? > > > > The rule by which the diagonal is constructed makes it differ from both > > representations when those representations have equal values. > > The rule does not make 1.000... differ from 0.999.... Why should it? The rule makes the diagonal different from either, and from any decimal containing a 0 or 9 digit in any position whatsoever. > > > > > In any case, for the set of all binary sequences (not numbers) no two > > are the same, so the issue does not arise > > I am interested in the number of real numbers only. There are at least as many binary sequences as reals in [0,1], as a simple surjection from binaries to real shows. |