Cantor Confusion
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From: William Hughes on 27 Jan 2007 12:38 On Jan 27, 11:58 am, mueck...(a)rz.fh-augsburg.de wrote: > On 27 Jan., 17:38, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Jan 27, 8:33 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > > > > > Induction can possibly > > > > prove that all the members of V* have some property, but can prove > > > > nothing about V* itself. > > > We can boil down the discussion about trees to the following simple > > > question, considering only one path, for instance the path p on the > > > outmost left hand side of the tree. This path p (in terms of nodes) is > > > the union of all paths of finite trees with length n, n in N. > > > Therefore all the path-*lengths* in the union are natural numbers. > > > Notwithstanding the question whether there are infinitely many paths > > > in the union or not: If the union path p is infinite, then at least > > > one of the paths in the union must be infinite. > > > > Is this so? > > No. > > > Think of the EIT. The diagonal is the union of the lines. > > None of the lines is an (potentially) infinite set. > > The diagonal is an (potentially) infinite set. > > > A union of finite sets can be (potentially) infinite. >Fine. (Please use "numbers". The paths-lengths are numbers (of a > unit). ) But set theorists deny this. They say: A union of finite > numbers cannot yield an infinite number. No. They say exactly the opposite. They say the union of any unbounded set of natural numbers (e.g. all natural numbers) is an infinite number. > So a union of different > natural numbers must contain an infinite number should it exíst? > > > > > The fact that p is a union of finite paths does not tell > > us whether p is finite or (potentially) infinite. > Fact is that p has no upper bound. Ok? So p does not have length n for any finite natural number n. So p does not have a finite path length. Either we say that p does not have a path length, or we say that the path length of p is an infinite number. - William Hughes
From: Carsten Schultz on 27 Jan 2007 12:45 mueckenh(a)rz.fh-augsburg.de schrieb: > > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > >> Induction can possibly >> prove that all the members of V* have some property, but can prove >> nothing about V* itself. > > We can boil down the discussion about trees to the following simple > question, considering only one path, for instance the path p on the > outmost left hand side of the tree. This path p (in terms of nodes) is > the union of all paths of finite trees with length n, n in N. > Therefore all the path-*lengths* in the union are natural numbers. > Notwithstanding the question whether there are infinitely many paths > in the union or not: If the union path p is infinite, then at least > one of the paths in the union must be infinite. Again one of WM's old `arguments': An infinite set of natural numbers must contain an infinite number. There cannot be an infinite set of finite natural numbers. > Is this so? :-) -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: Virgil on 27 Jan 2007 14:10 In article <1169904834.634452.44410(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > > > Induction can possibly > > prove that all the members of V* have some property, but can prove > > nothing about V* itself. > > We can boil down the discussion about trees to the following simple > question, considering only one path, for instance the path p on the > outmost left hand side of the tree. This path p (in terms of nodes) is > the union of all paths of finite trees with length n, n in N. > Therefore all the path-*lengths* in the union are natural numbers. > Notwithstanding the question whether there are infinitely many paths > in the union or not: If the union path p is infinite, then at least > one of the paths in the union must be infinite. Only in WM's weirdsville is this idiocy true. In the set of unary trees (naturals), the union of all finite naturals is not finite but does not contain anything but finite members, so why should any union of finite trees work any differently? > > Is this so? In fact we have just seen an example, the naturals, for which it is false that a union of finite sets of finite objects need contain a non-finite object to have a non-finite union. > > > > All I require is that a property holds (or not) for all the natural > > > numbers or all the levels or nodes or edges of trees which can be > > > enumerated by natural numbers. > > > Then you best conclusion is that it holds for all natural numbers but > > not that it holds for N. > > Precisely that is what I want and what I do! So how does that make the union of all finite naturals contain as a member any infinite natural in order to be not finite?
From: Virgil on 27 Jan 2007 14:15 In article <1169905223.354212.199580(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 26 Jan., 02:21, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote:> > A > path-length which has no upper bound may be sufficent to be called > > > infinite. > > > > Sorry, a path-length is something fixed, so the wording is inadequate. > > But if you call a path-length infinite you do not satisty (1), because > > there you associate path-lengths with natural numbers. > > We can boil down the discussion about trees to the following simple > question, considering only one path, for instance the path p on the > outmost left hand side of the tree. This path p (in terms of nodes) is > the union of all paths of finite trees with length n, n in N. > Therefore all the path-*lengths* in the union are natural numbers. > Notwithstanding the question whether there are infinitely many paths > in the union or not: If the union path p is infinite, then at least > one of the paths in the union must be infinite. This is just WM claiming once more that if the union of all naturals is a set that is not finite then there must be a natural number in one of those finite sets being unioned which was already not finite. Wm has this persistent delusion that a union of sets can contain objects not in any of the sets being unioned. > > Is this so? No, it is not so!
From: Virgil on 27 Jan 2007 14:18
In article <1169910030.783753.50380(a)a75g2000cwd.googlegroups.com>, davidmarcus(a)alum.mit.edu wrote: > On Jan 27, 8:33 am, mueck...(a)rz.fh-augsburg.de wrote: > > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > > > > > Induction can possibly > > > prove that all the members of V* have some property, but can prove > > > nothing about V* itself. > > > We can boil down the discussion about trees to the following simple > > question, considering only one path, for instance the path p on the > > outmost left hand side of the tree. This path p (in terms of nodes) is > > the union of all paths of finite trees with length n, n in N. > > Therefore all the path-*lengths* in the union are natural numbers. > > Notwithstanding the question whether there are infinitely many paths > > in the union or not: If the union path p is infinite, then at least > > one of the paths in the union must be infinite. > > > > Is this so? > > The way to answer questions in mathematics is to provide a proof. Can > you prove the claim or its negation? As WM is declaiming that a union can, and in some circumstances must, contain an object not a member of any of the sets being unioned, a proof would be quite difficult. Perhaps that is why WM never tries to prove his claim. |