Cantor Confusion
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From: mueckenh on 20 Dec 2006 06:30 Newberry schrieb: > >The edges have cardinal number aleph_0. > > Certainly. > > > A > > subset of the paths can be shown to be in bijection with the real > > numbers of the interval [0,1]. > > Which subset? Path as a set of intervals or path as a set of eges? How > does the above follow from this? 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. We have aleph_0 = |{edges}| >= |{paths}| >= |{reals of [0,1]}| = 2^aleph_0 while according to set theory aleph_0 < 2^aleph_0 Regards, WM
From: mueckenh on 20 Dec 2006 06:36 William Hughes schrieb: > > > No. Only properties of the diagonal that you > > > have accepted are used. > > > The conclusion is that there is no line. > > > > Every element of the diagonal is n some line. > > Every element of the diagonal is a finite natural number, by > > definition. > > > > Up to the finite natural number n: Every element =< n of the diagonal > > is contained in line n. Induction shows this is valid for any finite > > natural number. > > > > There are only finite numbers in the diagonal. QED. > > Given that the diagonal is the potentially infinite > set of natural numbers, the fact that there are > only fiinite numbers in the diagonal is immediate. > The only property of the diagonal that is used is > the fact that it is not bounded. In particular > > -it is assumed that the diagonal contains > only finite numbers > > -it is not assumed that all the elments of the > diagonal exist > > The contradiction holds, using only properties > of the potentially infinite set of natural numbers that > you have stated. It does not as I have written in my last letter to you. But even if your proof would hold, then we would have the fact that two contradictory results can be derived. Regards, WM
From: mueckenh on 20 Dec 2006 06:45 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. Regards, WM
From: mueckenh on 20 Dec 2006 06:54 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. 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. Regards, WM Regards, WM
From: mueckenh on 20 Dec 2006 06:58
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. > > 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. Regards, WM |