Cantor Confusion
in [Math]
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From: Franziska Neugebauer on 12 Feb 2007 08:30 mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Feb., 13:53, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: [...] >> > Please use the coordinate system which I have defined. >> >> Reducing a tree or of a path to a simple set of nodes is accompanied >> with a loss of information, namely the loss of the information of the >> connectedness of the nodes. > > The connectedness is given by the stucture of the tree. Your claim is: | >One cannot determine merely from a set of nodes for a given | >tree whether that set of nodes is or is not a path. [<-- FN] | One can. [<-- WM] As I have shown in my example one cannot. How long do you want to argue that fact? > The connectedness of elements of a matrix is given by the structure of > the matrix. 1. I don't see any relevance to your claim. 2. Please define structure of a matrix. 3. Since when are elements of a matrix "connected"? Please define: "To elements m_ij and m_kl of a matrix m are connected iff, ..." [...] >> Hence your "coordinate system" does not >> allow binary trees to be "located" at a unique position. In other >> words: A set of nodes is not sufficient to uniquely *determine* trees >> or paths. > > No, but it is sufficient to determine uniquely all paths of my tree. > See the simple example above. If the object of your investigation is a special, single, hand-crafted, unique tree you should try to avoid positing claims which merely sound universally valid. Fallacy of faulty generalization. >> As I have shown in my example solely from the set of nodes of a tree >> and a set of nodes of a path you cannot generally decide whether >> these sets have been originally derived from a tree which contains >> that path or does not contain it. > > Given my fixed cordinate system, there remains no ambiguity at all. Then we are no longer talking about binary trees in general. F. N. -- xyz
From: mueckenh on 12 Feb 2007 08:48 On 12 Feb., 13:59, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > Set theory, however, says: If the union of all finite paths, > > 0., 0.0, 0.00, 0.000, ... , > > is in the tree, then the infinite path p(oo) = 0.000.... is in the > > tree too. > > No contemporary set theory or graph theory does posit such a nonsensical > claim. It is entirely *your* claim, better known as It is easy to recognize from the union of all finite segments of N. Set theory says: If the union of all finite segments, {1,2,3,...,n} is present, then the infinite set N = {1,2,3,...} is present. (In fact set theory says: N is the union. But I don't know how this is accomplished.) Or do you see a difference betwee n the union of all finite segments and N? Regards, WM
From: mueckenh on 12 Feb 2007 08:59 On 12 Feb., 14:29, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > > The connectedness is given by the stucture of the tree. > > Your claim is: > > | >One cannot determine merely from a set of nodes for a given > | >tree whether that set of nodes is or is not a path. [<-- FN] > | One can. [<-- WM] > > As I have shown in my example one cannot. How long do you want to argue > against facts? All you have shown is that some clumsy notation is always capable of destroying meaning. > > > The connectedness of elements of a matrix is given by the structure of > > the matrix. > > 1. I don't see any relevance to your claim. Sorry, to make it easier seems impossible to me. > > Given my fixed coordinate system, there remains no ambiguity at all. > > Then we are no longer talking about binary trees in general. I never did so. I use one very special tree and its finite subtrees as a tool for a general proof. (I like trees in nature, like the trees in the Harz mountains, but I have no special connection to trees of graph theory.) Regards, WM
From: G. Frege on 12 Feb 2007 08:59 On 12 Feb 2007 05:48:47 -0800, mueckenh(a)rz.fh-augsburg.de wrote: > > Set theory says: If the union of all finite segments, {1,2,3,...,n} > is present, then the infinite set N = {1,2,3,...} is present. > Only in WM's dream world. (Hint: "is present" is not a term used in any standard set theory.) > > In fact set theory says: N is the union. > That's better now. > > But I don't know how this is accomplished. > It's "accomplished" the following way: N = U {1,...,n}. (*) neN With other words, (*) is a theorem in, say, ZFC. You can also formulate it the following way: N = U {{1,...,n} | n e N}, (**) i.e. the union of the set {{1}, {1,2}, {1,2,3}, ...} (containing all finite segments of the form {1,...,n} with n e N as elements) is N. F. "Learn some logic. Learn some mathematics. Or better yet, give up mathematics and take up basket weaving." (quote from sci.math, @WM) -- E-mail: info<at>simple-line<dot>de
From: G. Frege on 12 Feb 2007 09:23
On Mon, 12 Feb 2007 14:59:42 +0100, G. Frege <nomail(a)invalid> wrote: > > It's "accomplished" the following way: > > N = U {1,...,n}. (*) > neN > > With other words, (*) is a theorem in, say, ZFC. > > You can also formulate it the following way: > > N = U {{1,...,n} | n e N}, (**) > > i.e. the union of the set {{1}, {1,2}, {1,2,3}, ...} (containing all > finite segments of the form {1,...,n} with n e N as elements) is N. > We even have: N = U {n}. neN And/or N = U {{n} | n e N}, i.e. the union of the set {{1}, {2}, {3}, ...} (containing all singletons {n} with n e N as elements) is N. With other words, U {{1}, {2}, {3}, ...} = {1, 2, 3, ...}. F. "Learn some logic. Learn some mathematics. [...]" (a quote from sci.math, @WM) - But I doubt, you ever will. :-) -- E-mail: info<at>simple-line<dot>de |