Cantor Confusion
in [Math]
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From: WM on 14 May 2007 09:48 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? Regards, WM
From: WM on 14 May 2007 09:54 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. > > > 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. Regards, WM
From: WM on 14 May 2007 10:04 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. 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. All other paths of your set are waste. Regards, WM
From: WM on 14 May 2007 10:08 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? Then all the paths of the CIBT are countable. Fine. Regrads, WM
From: William Hughes on 14 May 2007 11:11
On May 14, 9:48 am, WM <mueck...(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? 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. > 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. 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 . |