Cantor Confusion
in [Math]
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From: Virgil on 7 May 2007 12:55 In article <1178538334.749731.40740(a)y80g2000hsf.googlegroups.com>, WM <mueckenh(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. He has certainly managed to point out one of WM's foibles.
From: Aatu Koskensilta on 7 May 2007 13:24 On 2007-05-07, in sci.math, Virgil wrote: > Some definitions of pi are infinite but quite well understood. Really? What definitions would these be? -- Aatu Koskensilta (aatu.koskensilta(a)xortec.fi) "Wovon man nicht sprechen kann, daruber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: WM on 7 May 2007 15:19 On 7 Mai, 17:46, Virgil <vir...(a)comcast.net> wrote: > In article <1178532990.891073.253...(a)y5g2000hsa.googlegroups.com>, > > WM <mueck...(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? > > You are the one pushing "finitely definable", whatever that means. That means: You cannot determine or define or construct or write down or tell me a bijection between the set of numbers which are definied by a definition identifying them uniquely and the set of natural numbers. > > But as the construction of the Cantor diagonal is finitely defineable, > and finitely defined, there should be no objection to it. That is not the point here. The point is that there are subsets of countable sets which do not allow for a bijection with N. Translate this fact to the binary tree. Regards, WM
From: WM on 7 May 2007 15:26 On 7 Mai, 18:06, Virgil <vir...(a)comcast.net> wrote: > In article <1178533278.892846.274...(a)w5g2000hsg.googlegroups.com>, > > > The number of separated paths is less 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. Yes. But if you consider two infinite sequences, namely the sequence 1,2,3... and the sequence 2,4,6,..., and put the tems in bijection, 1-2, 2-4, 3-6, ..., then you can wait and wait and wait: The first term will never get larger than the second. Even in the infinite this will not happen. Nevertheless you wait and wait and, look there is a pink William, the sum SUM[n= 0 to oo] (2-1-1) gets uncountable. It is a pity. Regards, WM
From: Virgil on 7 May 2007 15:29
In article <1178535380.443355.198500(a)q75g2000hsh.googlegroups.com>, William Hughes <wpihughes(a)hotmail.com> wrote: > On May 7, 6:23 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 6 Mai, 22:26, Virgil <vir...(a)comcast.net> wrote: > > > > > In article <1178461785.380972.277...(a)e65g2000hsc.googlegroups.com>, > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 6 Mai, 03:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > In article <1178190762.081101.159...(a)h2g2000hsg.googlegroups.com> WM > > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > > > Wrong. If there is always a path p' with p, > > > > > > > What does this statement mean? > > > > > > Even in the *infinite* tree a path cannot be distinguished from all > > > > other paths. That means, a real number which cannot be described by a > > > > finite formula (and most of them cannot) does not exist. > > > > > What axiom requires that a number must be defined by a finite formula in > > > order to exist? > > > > What kind of existence has an undefinable number? Axiomatic? > > > > > > > > > > Cantor's diagonal proof fails, because the diagonal number is never > > > > distinguished from all other real numbers (if uncountably many real > > > > numbers exist). > > > > > Cantor's "diagonal" proof only involves binary strings, not numbers, and > > > a very simple rule distinguishes the "diagonal" form every member of the > > > list. > > > > Only finite strings have been tested so far. > > > > Regards, WM > > > > There are an infinite number of members to check. However, each of > them > only requires checking a finite string. > > Whoops > > Over there! A pink elephant! > > There are an infinite number of members to check. So to check > them all you have to check an infinite string. > > - William Hughes |