Cantor Confusion
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From: Virgil on 19 Jan 2007 15:25 In article <1169214359.067976.72630(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > You shall once and for all commit yourself to whether your want to > > argument with > > > > 1. Union of trees. > > 2. Union of nodes. > > 3. Union of edged. > > 4. Union of paths. > > > No. Each one will be applied where approriate. Otherwise you cannot see > the contradiction. The contradiction is in WM's false assumption that nodes and edges are eniugh. But unless (4) is applied and can be shown to generate ALL infinite paths, one cannot guarantee a complete infinite binary tree by any such unioning. For infinite trees, nodes and edges are not enough to determine a tree uniquely, as there are different infinite trees which have the same set of nodes and the same set of edges, but different sets of paths. For example: The infinite binary tree of eventually constant paths (from some node onward all branches of a path are in the same direction) has the same set of nodes and the same set of edges but not the same set of paths as the complete infinite binary tree.
From: Virgil on 19 Jan 2007 15:37 In article <1169214789.690717.101110(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169111252.322725.263330(a)a75g2000cwd.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > T1, the union of all finite trees, contains only finite paths. > > > > > > It would be nice if you could show the existence of at least one finite > > > path in this union. Can you? Where does a path end? > > > > With its last (terminal or leaf) node, which in a finite tree every path > > has. > > But which node is that if we have the infinite union? Every node of a > path is a terminal node in some tree. But the paths in the union simply > refuse to stop on Virgil's command! In eternity! Unless WM defines his infinite paths as infinite sets or seqeunces of finite paths, he will have no infinite paths in his "union" of trees. And if he does define them that way, he has, essentially, as many infinite paths as there are infinite sets of finite paths. And since there are countably many finite paths there are uncountably many infinite sets of finite paths. Alternately, on can view an infinite path as an infinite sequence left or right branchings, but this gives the set of all infinite sequences of left/right branchings which is again uncountable.
From: Virgil on 19 Jan 2007 15:39 In article <1169215035.451242.76530(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > A path is a subset of the set of nodes. > > > > Since every subset of the set of nodes of a tree is a set of nodes you > > eventually say: "A path is a set of nodes". This is not the whole truth. > > > > A path _has_ a set of nodes (besides the information about the > > connectivity of these nodes). > > > > > That does not mean that every subset of the set of nodes is a path. > > > > Actually no set of nodes _is_ a path. > > In a well defined tree, as my CT and my WWT are, every path is an > ordered set of nodes. It is a subset of the complete ordered set of > nodes which is the infinite tree. The sets are even well ordered. But there are infinitely many nodes so that the set of such infinite subsets of the set of nodes is uncountable.
From: Andy Smith on 19 Jan 2007 16:00 David Marcus writes > >You have demonstrated what people on sci.math refer to as dyslexia. You >have switched the order of the operations. As far as the limit is >concerned, x is a fixed number. Let's try x = 1/2. The question is what >is the value of > > lim n->oo (1/2)^n > >? To be more precise, saying > > lim n->oo |x|^n = 0, for |x| < 1 > >is different from saying > > lim n->oo sup_{|x|<1} |x|^n. > >"sup" is like max, but is used when the maximum isn't obtained. > But, if the formula is to be true for all x<1, then don't we need to show that lim n->oo sup_{|x|<1} |x|^n = 0 ? Which was what I was trying, unsuccessfully and amateurishly, to do/disprove? that is exactly the point at issue - there is no question that lim n->oo |x|^n = 0 for any |x| that is finitely different from 1. The issue is whether there are any wrinkles and whether the limit is true for all |x| strictly <1 i.e. all reals in the interval from 0,1 excluding the exact point at x = 1. -- Andy Smith
From: Andy Smith on 19 Jan 2007 16:14
David Marcus writes > >You have demonstrated what people on sci.math refer to as dyslexia. You >have switched the order of the operations. As far as the limit is >concerned, x is a fixed number. Let's try x = 1/2. The question is what >is the value of > > lim n->oo (1/2)^n > >? To be more precise, saying > > lim n->oo |x|^n = 0, for |x| < 1 > >is different from saying > > lim n->oo sup_{|x|<1} |x|^n. > >"sup" is like max, but is used when the maximum isn't obtained. > In this situation, if we want to show that |x|^n converges to zero for all |x| < 1 as n->oo, we can consider an eta>0, and ask, at a given value of n, what range of |x| satisfies |x|^n <eta. Then we can then ask, as n->oo whether we can reduce eta to 0, such that the range of |x| satisfying |x|^n <eta is |x|<1. Is that dyslexic? I saw this as different from a usual convergence problem, precisely because it has to be shown true for all |x| in [0,1). But maybe I am well adrift? -- Andy Smith |