Cantor Confusion
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From: Virgil on 20 Dec 2006 15:12 In article <1166613609.404263.250020(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > > > Element n+1 can be shown to exist in L_D (which is obviously > > > > > > > > > a line > > > > > > > > > containing n+1). > > > > > > > > > > > > > > > > No. L_D is bounded. The largest element of L_D is n. > > > > > > > > L_D does not contain n+1. > > > > > > > > > > > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > > > > > contained in the diagonal. If your L_D does not contain them, > > > > > > > then you > > > > > > > have the wrong L_D. > > > > > > > > > > > > Assume that there exists an L_D which contains all the numbers > > > > > > contained in the diagonal. L_D is bounded, > > > > > > > > > > If an unbounded diagonal exists, then obviously an unbounded line > > > > > must > > > > > exist. > > > > > The conclusion is false, so the antecedent cannot be true. > > > > > > > > The diagonal is the potentially infinite set of natural numbers. > > > > This exists and is unbounded. > > > > > > Yes, but not in the way you understand "to exist". Potentially infinite > > > means: always finite. The diagonal is always finiter though not > > > bounded > > > > The only property of the diagonal that is used is that > > it is unbounded (i.e. given a finite set of elements of the diagonal, > > it is possible to add one element). > > That property is correct. > > > The existence of all the elements > > of the diagonal is neither claimed nor used. The "infiniteness" > > of the diagonal is neither claimed nor used. > > Fine. > > > > The contradiction (a line must be both bounded and unbounded) > > No line must be unbounded in case of a potentially infinite (i.e. > always finite) diagonal. WM, in allowing for, and even requiring, the existence of an infinite binary tree, implies the existence in ZFC or NBG of a model of his EIT in which every line is finite with a last member but the diagonal is /actually/ infinite with no termination. Any of the infinite paths in such a tree is such a model, with any path as the diagonal and the sections of that path from root node to some other node as lines. This model satisfies all of the necessary criteria for WM's EIT, but show that WM's claims for such EIT's are false.
From: Virgil on 20 Dec 2006 15:23 In article <1166614207.528415.222250(a)i12g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > The paths are binary representations of all real numbers of the > interval [0,1]. Some rational reals have double representations, namely > such ending by 000... and such ending by 111... like 1.000... and > 0.111... . So there are not less real numbers in the interval [0, 1] > than paths in the binary tree. WM's sloppy logic again fails him! What WM has argued justifies "there are not less paths than reals", which is not what he is claiming, but not "there are not less reals than paths", which is what he is claiming. The claimed equality is validated by noting, in addition, that there are only countably many double representations but uncountably many paths and reals, so that a merely countable discrepancy may be ignored.
From: Virgil on 20 Dec 2006 15:31 In article <1166615108.048683.78660(a)t46g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > That does not prove (*) since the whole diagonal is _not_ bounded by any > > such n. > > If you doubt my proof, please give an example using a finite number n > eps N for which it does not hold. If you cannot find such a number, > remember that N contains only finite numbers. But infinitely many of them. Consider the model given within any infinite tree by any of its (infinite) paths. The path is the diagonal and each line is the section of that path from the root node to any other node. Each edge in the path is labeled by (identified with) the natural number of edges in the line of which it is the last edge. In this model, every natural appears in some line. In this model, there is no line in which every natural appears. This model disproves WM's contentions about his EIT.
From: Virgil on 20 Dec 2006 15:44 In article <1166615654.966391.63070(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > You have correctly show that: > > > > A: For every natural number n, there is a line L(n), such > > that n is contained in L(n). > > > > You need to prove > > > > B: There is a single line, L_D, such that every natural > > number n is contained in L_D. > > > > A and B are not the same statement. You have not given > > a proof of B. (B follows from (A + linear ordering) iff there > > is a last line) > > Please give a counter example for a finite number n. Consider the model given within any infinite tree by any of its (infinite) paths. The path is the diagonal and each line is the section of that path from the root node to any other node. Each edge in the path is labeled by (identified with) the natural number of edges in the line of which it is the last edge. In this model, every natural appears in some line. In this model, there is no line in which every natural appears. This model disproves WM's contentions about his EIT. > Remember, even if there is not a last line, every n eps N is finite. > Therefore your claim is void, unless you can find a linear set of > finite elements which forbids exchange of quantifiers. But spare me > with dancers etc. See the above model, built within one of these infinite binary trees which WM has already used in his arguments. In this model, it is true that "For every n in the diagonal, there is a line containing n" or in symbols "A n in D ( E L in EIT ( n in L )) but "It is false that there is a line containing every n in the diagonal" or, in symbols "~ (E L in EIT ( A n in D ( n in L))" In fact, more directly: For every line there is a natural in the diagonal not in that line. or A L in EIT (E n in D ( not n in L))
From: Virgil on 20 Dec 2006 15:48
In article <1166615903.717095.36630(a)f1g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > Because each element of the diagonal beongs to a line which contains > > > all peceding elements. > > > > WM is conflating things again. That each element of the diagonal must > > belong to SOME line does not mean that they all must belong to the same > > line. > > Show me two elements wich cannot belong to the same line. How is that in any way relevant? What I have said allows that for each pair there is SOME line which they do not both belong to. That line is the one which ends with the smaller element. > > > > Consider a new triangle each "line" containing only one element, what > > was in the former triangle the last element, and the diagonal containing > > all these elements. > > > > It is still true that every element in the diagonal is in some line, but > > now it is obvious that there is no line which contains all elements of > > the diagonal. > > Why do you think to need an example of this poor class? Argue within > the EIT. Consider the model of EIT given within any infinite tree by any of its (infinite) paths. The path is the diagonal and each line is the section of that path from the root node to any other node. Each edge in the path is labeled by (identified with) the natural number of edges in that line of which it is the last edge. In this model, every natural appears in some line. In this model, there is no line in which every natural appears. In this model, WM is wrong! This model disproves WM's contentions about his EIT. |