Cantor Confusion
in [Math]
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From: William Hughes on 14 May 2007 11:15 On May 14, 10:04 am, WM <mueck...(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 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... > You see that there was absolutely no reason to include them. If the > set of paths covers all infinitely many nodes of p, Look! Over There! A pink elephant. When you remove the second last path you are left with the last path. > then there is only > one path p* in the set which covers all these nodes, - William Hughes
From: Virgil on 14 May 2007 14:49 In article <1179150494.751572.133480(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Mai, 19:53, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 13, 12:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 13 Mai, 15:47, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > 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. > > > > > > The simplest way to do this is one path per node. > > > > > > 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? But why then should the tree contain the supremum of R? What is WM's definition of a "supremum" of R? The "natural" ordering on R is by number of nodes in common with p. This number is finite for every member of R, but the set of such numbers has no supremum in N. So if R is to have a supremum, at least by that natural ordering, it must be p, which is not a member of R. Or does WM have some other ordering of R in mind?
From: WM on 14 May 2007 15:17 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. Is p the supremum of the set of intersections with other paths p*, i.e. is p = sup {p n p* | p* e T \ {p}}? Is p the union of the set of intersections with other paths p*, i.e. is p = U{p n p* | p* e T \ {p}}? Is p the set p = {p n p* | p* e T \ {p}}? > > 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. Regards, WM
From: Virgil on 14 May 2007 15:21 In article <1179150859.473726.299240(a)e51g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Mai, 20:18, Virgil <vir...(a)comcast.net> wrote: > > In article <1179052221.849088.146...(a)l77g2000hsb.googlegroups.com>, > > > > A set of paths p* =/= p can be used include them all, i.e., all nodes > > > of p: > > > > > U{k(p, n) | n in N} c U{k(p*, n) | p* =/= p & n in N} . > > > > > This is only possible, if the right hand side union includes all nodes > > > of p. This is only possible if > > > it includes the path p. > > > > So that now WM claims to have a path p* containing every node of p and > > some other nodes as well? > > No, I claim that a set of paths, which cover all the nodes of p, > necessarily contains a path which is p. That claim is demonstrably false if the set of paths is allowed to be infinite. Given any path, p, i n a CIBT,mfor each n in N, let p_n be a path which coincides with p up to level n, but branches oppositely from p in going from level n to level n+1. then no such path is p itself. For every node in p there is such an n in N and for every n in N there is such a path, p_n, and none of these paths can be the same as path p, so the set of these paths (a) has at least one path containing any node of p and (b) does not contain the path p itself. > > > > > In order to establish existence of your alleged foreign path, you would > > > > have to prove, contrary to fact, that all the pieces of foreign paths > > > > which yu claim exist belong to the same foreign path, which they do > > > > not. > > > > > > > You do not oppose to (1). If you oppose (2) and claim that one > > > > > foreign > > > > > path does not suffer, then please prove that at least two foreign > > > > > paths are required. > > > > > > In fact, infinitely many foreign are required, but no one of them > > > > contains all of p. > > > > > If a set of infinitely many foreign paths can do it, then one foreign > > > path can do it. > > > > If any one path contains all the nodes of p, it cannot also contain any > > nodes not in p, so it must be p. > > Correct, therefore the condition (B) > A k (k in p -> E p* ((p* =/= p) & (k and all of its predecessors in > p*))) > is illusionary. Only in WM's illusionary world. In our world, each k is at some level n in N, so that one can take the p_n defined above as p*. So in our world condition (B) is actuality. WM's confusion looks a bit like like quantifier dyslexia. As (B) is stated, each p* depends on on which k was chosen. WM seems to think that there ought to be a single P* whichworks for all k, but that is not requires by (B).
From: WM on 14 May 2007 15:24
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, > > > When you remove the second last path you are left with > the last path. We need no last path in order to see that there is no set of nodes which cannot be covered by one single path. That has nothing to do with infinity. And, therefore, it cannot be remedied by infinity. Even the help of a pink elephant is not sufficient, to take the unnecessary paths for necessary paths and the insufficient paths for sufficient paths. Regards, WM |