Cantor Confusion
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From: Bob Kolker on 15 Dec 2006 07:10 Han de Bruijn wrote: > > Brings me to the question: who is doing Geometry these days? Yoo Hoo! Toplogical Manifold Theory is geometry in its sweetest most beautiful form. It is Pythagoras' and Euclid's dream realized. In particular we understand gravitation best as a geomtetric property of physical spacetime manifolds. As Platon said: God forever geometrizes. Platon was not only a wise guy, he was a wide guy. Bob Kolker
From: mueckenh on 15 Dec 2006 07:12 Virgil schrieb: > In article <1166092755.336596.309060(a)l12g2000cwl.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > WM deceives himself over the number of lines versus the number of > > > elements in any one line. The number of lines is not finite but the > > > number of elements in any one line is finite. > > > > Each line differs by 1 element from the preceding line. If the number > > of lines is actually infinite then the number of differences must be > > actually infinite too. > > So far so good. > Their sum is an infinite number. If all elements are there, then also all sums are there. If not all sums are there, not all elements are there. Regards, WM
From: Bob Kolker on 15 Dec 2006 07:14 Han de Bruijn wrote: > > > Suppose you mean "Is". What does it mean that a set is unbounded? It does not have a largest element in its ordering. The theory of ordinals has provided the well ordered set of natural integers with an least upper bound, to wit, omega, the first infinite ordinal. The integers by themselves are not bounded above. Given any integer one can find a large integer by adding one. Bob Kolker
From: mueckenh on 15 Dec 2006 07:17 Virgil schrieb: > But the set of all sequences of edges is NOT countable, so what > restrictions on sequences does WM have to limit then to only a countable > subset of all such seqeunces? The set of all edges is countable. The set of all sequences could be uncountable, if all combinations of edges were possible. But that is not the case in the binary tree. No path can separate from another one without an edge for each of them. > > > > Further, the tree is continuous in that at every level n there arrive > > 2^n edges, that means there arrive 2^n separated paths. > > WRONG! there "arrive" 2^n separated subsets of paths There arrive 2^n edges as well as 2^n separated (subsets of) paths. > with each subset as > numerous as the whole set. > It is useless to emphasize this simple fact because for every path of that subset there is another edge. > > If in the whole > > tree 2^aleph0 paths have been separated, then there must be a level up > > to which (or within which) less than 2^aleph0 paths have been > > separated > > One is not separating individual paths but sets of paths with each set > uncountable as the set of all paths. And that is the very real reason of your confused mathematics: There are no seprated paths at all. But you claim them uncountable. ======================= variouzs answers: > But built uncountably many sequences of them. I built all the sequences of them wich are paths in the tree. > > Any book on set theory including ordinals will be sufficient. > And will show that 2^omega is not countable. Your attitude is understandable but I had not expected it from you. And I am really intrigued about it. Obviouse absurdities of set theory which have not yet explictly been learned by its students seem to be recognizable as absurdities even by such hard boiled students like you. Learn: It is in fact funny that 2^omega is countable. But it is set theory. So you will be able and inclined to easily swallow it. (It would be better, of course, to take it as the nonsense which it is.) > The "matrix" is a fiction. Every irrational number is a fiction. > What have any of Cantor's proofs of > uncountability to do with trees? They are contradicted by the tree. > Sure it can. The union of a family of sets need not be a subset of any > of them. Not in general, but if the elements are linear ordered sets. > But unless we assume that there can only be finitely many we cannot > conclude that there is a last or largest conctaining all the others. It is enough to know that each one is finite. Regards, WM
From: mueckenh on 15 Dec 2006 07:21
Virgil schrieb: > > > > No I simply state that the diagonal proof fails in case of the tree. > > AS the diagonal proof is not allied to trees, so what? > Its contradiction is. > > > > We can simply count the edges. They are countable. So we an count the > > biginnings of separated parts of paths (because every edge is a > > beginnng of the separated part of a path, notwithstanding which it may > > be). > > > Every finite beginning of a set of paths, however long, is the beginning > of as many paths as the set of all paths. > > So there are as many paths following any such beginning as there are > paths altogether. And there are as many beginnings following such a beginning. > > > Therefore the beginnings of separated parts of the paths are > > countable. > > > It is the endings that are not countable. It is the endings, which do not exist. > > Do you deny that every edge is the beginning of that > > part of a path where it is separated from other paths? > > I deny that it has any relevance to the "number" of paths. To state it clearly: You have to believe (or at least to assert to believe) that there are more separated paths than beginnings of separated paths in the tree. ========================== > Without some assumptions (other than the rules of logic) one > cannot deduce anything about anything. > So one must start with SOME assumptions. No. One can start from some observations like 1+1 = 2. Regards, WM |