Cantor Confusion
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From: Virgil on 13 Dec 2006 17:37 In article <1166043658.530354.313800(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > mueckenh(a)rz.fh-augsburg.de schrieb: > KORREKTION > > A surjection onto the paths covers all elements of the *range* = every > path. > > Regards, WM AS the domain of that alleged surjection is the set of all endless sequnces of edges,or perhaps nodes, WM still needs to show that there is a surjection from the set of all edges (or nodes) to the set of all those sequences. A task that is quite beyond anyone, including WM.
From: Virgil on 13 Dec 2006 17:41 In article <1166043984.701298.29110(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > > > A_1 = {1} > > > > > > > > A_2 = {1,2} > > > > > > > > A_3 = {1,2,3} > > > > > > > > > > > > > > > > B = {1,2,3} > > > > > > > > > > > > > > > > then B is contained in the last A_i. If there is no last A_I, > > > > > > > > then > > > > > > > > there is > > > > > > > > no A_i that contains B > > > > > > > > > > > > > > That has nothing to do with "last". > > > > > > > > > > > > If A_i contains B, then A_i contains any A_j. > > > > > > Therefore A_i is "last". > > > > > > > > > > > > > > No comment? > > > > > > If B contains any A_i then B is the last. > > > > B cannot be the last A_i, as B is not > > one of the A's (B corresponds to the diagnonal, > > the A_i correspond to the lines) > > In which element does it differ from every A_i? In the (i+1)st element. Actually B differs from every A_i by HAVING an (i+1)st element, which an A_i does not. > Don't forget, the A_i > form a linear set. It is not possible for B to differ from A_n in > element a but not in element b and from A_m in element b but not in > element a. Irrelevant when: For every A_i, B has an element not in that A_i.
From: Virgil on 13 Dec 2006 17:51 In article <1166044694.629951.178340(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > You are aware of the number 2^omega being a countable one? If so, why > then do you speak of an uncountable set of shares? Because 2^omega, representing the set of functions from omega to {0,1}, is demonstrably not countable. > > Do you think that lim {n-->oo}1/2^n * 2^n consists of uncountably many > parts? I have no idea why WM thinks that relevant to anything under discussion, but it certainly does not effect the demonstrable uncountability of 2^omega. > > How can 1 + 1/2 + 1/4 + ... +1/2^n + ... yield the value 2 if we add > only countably many shares which are uncountably small? I have no idea why WM thinks that relevant to anything under discussion, but it certainly does not effect the demonstrable uncountability of 2^omega. > > Would you mind to think about that topic again? > > Now let's argue with aleph0 (in modern set theory it is the same as > omega, although there is a big difference). Aleph_0 is a cardinality. Omega is an ordinal. It depends on what definition one uses for cardinality whether they are the same set or not, and they are not always the same set. > > The problem with Cantor's theorem aleph0 < 2^aleph0 is that the > function f(x) = 2^x must be discontinuous What is WM's definition of "continuity"? It certainly cannot be any of the standard ones! So absent a definition, WM's claim is mere nonsense.
From: Virgil on 13 Dec 2006 18:00 In article <1166045232.670658.72540(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > > > > > > > > That one which goes left from the root is not engaged, because > > > > > > > we need > > > > > > > only half of the set of edges. > > > > > > > > > > > > So your construction is not a surjection. > > > > > > > > > > Of course it is! It is a surjection from the set of edges onto the > > > > > set > > > > > of paths. > > > > > > > > Than again. To what number maps the edge that goes left from the root? > > > > If it is a surjection, there should be one. > > > > > > > A surjection onto the paths covers all elements of the *range* = every > > > edge. > > > > What? The *range* of a surjection from edges onto paths is /not/ every > > edge; it is every path. The /domain/ of the surjection is every edge. > > (You actually teach Analysis, and you don't know this?!?!) > > Did you never copy and paste a wrong piece? > I corrected it already: The range is "every path". > > > > > It is not necessary that every edge is mapped on a path, to show > > > that there are not less edges than paths. It is only necessary that > > > there is no path without edge mapped on it. > > > > > > > That is poorly stated. The mapping that maps the first edge to the left > > to every path provides us with a mapping such that "there is no path > > without edge mapped to it"; but this mapping does not prove that the > > cardinality of the set of paths <= the cardinality ({a single edge}) = > > 1. > > You misunderstood. What I explained to Dik is another mapping than that > discussed by us. It is a mapping of one single edge on the path > representing 1/3, the next edge is mapped on the path representing pi, > and so on. I claim that for every real number I can name an edge to be > mapped on this real number. And noboy can disprove it by constructing a > diagonal number, because the tree contains them all. But Wm cannot do it with a single mapping. Each path can be uniquely represented by an infinite sequence of left or right branchings from the root node through all its successive nodes. Each edge can be uniquely represented by a finite sequence of left or right branchings from the root node to the end node of that edge. Despite his many claims, Wm has not shown any mapping which surjects the above set of finite sequences to the above set of infinite sequences.
From: Virgil on 13 Dec 2006 18:03
In article <1166045930.856918.153600(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > A function can be abbreviated as number variable or as variable > > > number. > > > > Could you explain, what "to abbreviate" means in the present context? > > > > Yes, but I'm not inclined to do it. > > Good night. WM is never inclined to explain what he cannot satisfactorily explain. All his claims involve hidden assumptions which he is "not inclined to explain" because he can't without revealing some of those hidden assumptions. |