Cantor Confusion
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From: Carsten Schultz on 18 Jan 2007 16:27 Virgil schrieb: > Each path in the complete infinite binary tree is essentially a > function from N to {0,1}, i.e., an infinite binary sequence of left(0) > or right(1) branchings. There is a level at which WM knows this. > But Cantor's diagonal proof, as originally stated, proved that the set > of all such sequences not to be countable. And to WM this is obviously wrong. Now he is trying to find a way to let others see this too. Of course, his arguments are never valid. But he is looking for a way to state the ever same wrong arguments that would convince others. There must be one, because to him it seems all so clear! Momentarily his tool of choice seems to be this binary tree. > So if WM's tree has only countably many paths, it must be leaving most > paths out. > > An several people have presented analyses showing where WM's omissions > occur. Of course. The real sad thing is that he teaches mathematics courses. Therefore while making a fool out of himself here, he also disgraces his school. Carsten -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: Virgil on 18 Jan 2007 16:33 In article <1169119020.460630.259250(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Carsten Schultz schrieb: > > > > You assume that the union P_i of paths contains more paths than can be > > > constructed from finite initial segments? > > > > I do not assume anything. I just note that being a path in the union of > > the T_i and being an element of the union of the P_i are a priori > > different things and that you would have to prove their equivalence in > > your setting should you claim this equivalence. > > > The union of all finite trees is an infinite tree. Given WM's special definition of union. > The countable union of all finite paths is in the union of all finite > trees. I am not at all sure what WM means by "The countable union of all finite paths". Unions are supposed to be of sets > The "complete" tree containing all paths is identical to the union of > al finite trees, with respect to nodes and edges. Since nodes and edges are not enough to determine an infinite tree, WM is ambiguous. Note that the incomplete infinite binary tree whose paths are all eventually constant, IIBT, and the complete infinite binary tree, CIBT, have the same set of nodes and the same set of edges, but one has a countable set of paths and the other an uncountable set of paths. So that nodes and edges alone only work for finite trees. For infinite trees, you need to look at the set of paths as well, and just edges and nodes don't give you that. As usual. WM has not done his homework before publishing, and so find himself in the wrong again. > Identical trees cannot contain different sets of paths. Infinite resse with identical sets of nodes and edges can! > Therefore, both trees contain the same set of paths. WRONG!
From: Virgil on 18 Jan 2007 16:38 In article <eooonv$ed3$1(a)news2.open-news-network.org>, Carsten Schultz <carsten(a)codimi.de> wrote: > Virgil schrieb: > > Each path in the complete infinite binary tree is essentially a > > function from N to {0,1}, i.e., an infinite binary sequence of left(0) > > or right(1) branchings. > > There is a level at which WM knows this. > > > But Cantor's diagonal proof, as originally stated, proved that the set > > of all such sequences not to be countable. > > And to WM this is obviously wrong. Now he is trying to find a way to > let others see this too. Of course, his arguments are never valid. But > he is looking for a way to state the ever same wrong arguments that > would convince others. There must be one, because to him it seems all > so clear! Momentarily his tool of choice seems to be this binary tree. > > > So if WM's tree has only countably many paths, it must be leaving most > > paths out. > > > > An several people have presented analyses showing where WM's omissions > > occur. > > Of course. > > The real sad thing is that he teaches mathematics courses. Therefore > while making a fool out of himself here, he also disgraces his school. And injures his students.
From: David Marcus on 18 Jan 2007 17:08 Carsten Schultz wrote: > The real sad thing is that he teaches mathematics courses. Therefore > while making a fool out of himself here, he also disgraces his school. Why does the school permit this? -- David Marcus
From: Michael Press on 19 Jan 2007 04:03
In article <MPG.2018a7a73eedebe1989b6d(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Michael Press wrote: > > In article <MPG.201731bcec6d9f73989b5c(a)news.rcn.com>, > > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > > > > Ah, you are at least a hundred years behind the times. No, a function is > > > most definitely not a formula. A function is a rule which assigns, to > > > each of certain real numbers, some other real number. For example, the > > > rule that assigns to each number a the number 0 if a is irrational and > > > the number 1 if a is rational is a function, but you will have a hard > > > time coming up with a formula (nor is a formula required). > > > > Hello. I agree. At least one-hundred-fifty years. > > Is this a formula? > > > > lim_{m -> oo} lim_{n -> oo} [cos (m! * x * pi)]^{2 * n} > > These days or in Euler's day? Both if you please. I have not seen formula precisely defined, so I am asking how it is typically thought of, as in your usage above. > Notice I said a "hard time". I didn't say it was impossible. I never thought you did, and agree that Andy Smith would have a _very_ hard time. -- Michael Press |