Cantor Confusion
in [Math]
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From: Virgil on 14 May 2007 15:27 In article <1179151441.352685.176080(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Mai, 20:43, Virgil <vir...(a)comcast.net> wrote: > > In article <1179062054.712048.154...(a)u30g2000hsc.googlegroups.com>, > > > > A very fine trick! But you > > > forgot to say why an infinite set of paths should be able to cover all > > > nodes of p. There is absolutely no reason. > > > > There absolutely is a reason! > > > > But it is so trivial that we did not think it necessary to state it. > > If WM's powers are too weak to figure it out for himself, then: > > > > Given any path p in a CIBT. > > For each n in N, Let p_n be a path that has in common with p > > the first n nodes of p, but no others. There is always such a path, > > any of the ones branching the other way at level n. > > Since every node of p is at some finite level n in N, the infinite > > set {p_n: n in N} of paths "covers" all nodes of p. > > > > Is that simple enough for you to grasp, WM? > > Now remove all the path p_1, and remove the path p_2, and remove .... > You see that there was absolutely no reason to include them. Then Wm is claiming that the empty set has a path in it that contains all the nodes of p, since p is specifically excluded from my set of paths. > If the > set of paths covers all infinitely many nodes of p, then there is only > one path p* in the set which covers all these nodes, namely p* = p. Wrong! There is no p in my set, in which every p_n differs from p at level n+1. > All other paths of your set are waste. Not if one wishes to exclude p itself, which is the whole point. It is WM's misunderstanding that is the waste.
From: Virgil on 14 May 2007 15:57 In article <1179151730.410966.154580(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Mai, 20:50, Virgil <vir...(a)comcast.net> wrote: > > In article <1179073448.864312.305...(a)h2g2000hsg.googlegroups.com>, > > > > > Therefore there is no set of paths p* the nodes of which form a > > > superset of the nodes of p. > > > > Actually every set of paths which has p as a member does that. > > Correct. > > > > But there are even sets of paths which do not contain p which do it. > > Incorrect. > > > > For example. for each n in N there is a path which coincides with p up > > through level n but at level n branches oppositely. Call it p_n. > > Clearly for each n in n, p_n =/= p. > > Also clearly, every node of p is in some p_n. > > So you now agree, that the set of all finite trees can be unioned (or > united?) to get the CIBT with not a single infinite path? No. Why should I agree to something so clearly false? How does WM manage to come to so many false conclusions? I do not agree with any of them. I have merely shown that there is a set of paths which does not contain, (and is to equal to) any given path in a CIBT but still manages to use every node of the given path. Every path in a CIBT is infinite, and the 'number' of such paths is beyond counting. > Then all the > paths of the CIBT are countable. Each path is of countable "length", but the set of such paths is as uncountable as the power set of N, with which it bijects.
From: William Hughes on 14 May 2007 16:09 On May 14, 3:17 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 14 Mai, 17:11, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 14, 9:48 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > For every node kn, of p, there is a path not equal to p > > > > > > which contains kn, call this p(kn). > > > > > > Let R be the set of all paths p(kn). > > > > > > R is an infinite set of paths that covers all the nodes of p. > > > > > > No. The infinite set R cannot be used because its supremum is p and, > > > > > therefore, not a path different form p. > > > > > We need a set for which every element is different from p. > > > > > Every element of R is different from p. > > > > > The supremum of R is not different from p. > > > > > No contradiction. R does not contain the supremum of R. > > > > No? > > > No. Every element of R contains a 1. The supremum of R > > does not contain a 1. R does not contain the supremum of R. > > So R does not cover every node of p. There remain infinitely many > nodes of p uncovered by every element of R. And as R does not contain > its supremum, even by the union of all elements of R not all nodes of > p are covered. > > > > > > But why then should the tree contain the supremum of R? > > > Stop trying to change the subject. The question is whether > > R is the set you said was impossible to find. > > It is important first to get clear about the tree if we discuss its > paths. However, we are discussing the paths in R. R is not a tree. <Attempt to change the subject snipped> > > > Which of the following statements is not true > > > Every element of R is different from p. > > > Every node of p is in some element of R > > Again only one side of the medal. <Attempt to change the subject > > Is the following statement also > true? > Look! Over there A pink elephant! > Most (more than half of all) nodes of p are outside of every union of > elements of R. > False. Most (more than half of all) nodes of p are outside of every *finite* union of elements of R. Back to the question of whether R is the set you said was impossible to find. Which of the following statements is not true Every element of R is different from p. Every node of p is in some element of R - William Hughes
From: William Hughes on 14 May 2007 17:28 On May 14, 3:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 14 Mai, 17:15, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > Given any path p in a CIBT. > > > > For each n in N, Let p_n be a path that has in common with p > > > > the first n nodes of p, but no others. There is always such a path, > > > > any of the ones branching the other way at level n. > > > > Since every node of p is at some finite level n in N, the infinite > > > > set {p_n: n in N} of paths "covers" all nodes of p. > > > > > Is that simple enough for you to grasp, WM? > > > > Now remove all the path p_1, > > > and you are left with infinitely many paths > > > >and remove the path p_2, > > > and you are left with infinitely many paths > > > > and remove .... > > > and you are left with infinitely many paths... > > more than half of which are useless for our purpose. > > > > > > You see that there was absolutely no reason to include them. If the > > > set of paths covers all infinitely many nodes of p, <silent internal snipping> This subthread has ended. - William Hughes
From: Virgil on 14 May 2007 19:22
In article <1179170250.996327.22270(a)k79g2000hse.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 14 Mai, 17:11, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 14, 9:48 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > For every node kn, of p, there is a path not equal to p > > > > > > which contains kn, call this p(kn). > > > > > > Let R be the set of all paths p(kn). > > > > > > R is an infinite set of paths that covers all the nodes of p. > > > > > > > No. The infinite set R cannot be used because its supremum is p and, > > > > > therefore, not a path different form p. > > > > > > We need a set for which every element is different from p. > > > > > > Every element of R is different from p. > > > > > > The supremum of R is not different from p. > > > > > > No contradiction. R does not contain the supremum of R. > > > > > No? > > > > No. Every element of R contains a 1. The supremum of R > > does not contain a 1. R does not contain the supremum of R. > > So R does not cover every node of p. Every node of p is a 0 and R covers every ode of p whichis a 0, so it covers all of p. > There remain infinitely many > nodes of p uncovered by every element of R. It is irrelevant that some member of R does not cover some node as long as another member of R does, and for every node, there is a member of R covering it. And not all of WM's double talk can cover up this failure of his claims. > And as R does not contain > its supremum That is the whole point! > even by the union of all elements of R not all nodes of > p are covered. If WM insists that some nodes are not covered, let him show us one of them. > > > > > > But why then should the tree contain the supremum of R? > > > > Stop trying to change the subject. The question is whether > > R is the set you said was impossible to find. > > It is important first to get clear about the tree if we discuss its > paths. It is important to WM to obscure the clear proof by construction that he is once again wrong. But he cannot do it without first proving that 2 = 1. > > > > Which of the following statements is not true > > > > Every element of R is different from p. > > > > Every node of p is in some element of R > > Again only one side of the medal. Is the following statement also > true? > > Most (more than half of all) nodes of p are outside of every union of > elements of R. Relevance? The only issue is whether, given a path p in a CIBT, there exists a set R of paths in that CIBT each distinct from p which contain collectively every node of p. William Hughes has constructed such a set. Others have also constructed such sets. They abound. WM has brought in many straw men in the vain hope of derailing the discussion from the issue, but has yet to deal with the issue itself. |