Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Virgil on 19 Dec 2006 23:06 In article <1166529472.820108.221940(a)t46g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166455723.089710.246540(a)f1g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > I don't need it. But there are others who do not see the forest because > > > of too many trees blocking the view. > > > > WM does need it to blind himself to the necessities of the situation. > > > > First WM denies infiniteness even in ZFC and NBG. > > Then WM discusses infinite binary trees, in which he divides an edge > > into countably many equal sized pieces. > > only in order to show that infinity yields contradictions. If there are infinite binary trees at all, which existence Wm has accepted and used as a basis for his arguments, then it is easily derived that every path is a model for a set of infinitely many finite naturals. Thus WM is hoist with his own petard.
From: Virgil on 19 Dec 2006 23:10 In article <1166529653.157491.94810(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > So you consider it nonsense that mathematics should consist of clear > > definitions and proofs of theorems. > > No. But I need no proof in order to see that > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > Absent some formal definition, "1 + 1/2 + 1/4 + ...+ 1/2^n + ..." is without any mathematical meaning whatsoever. And while you may think you "see" something, in mathematics it still requires proof.
From: Virgil on 19 Dec 2006 23:20 In article <1166530185.163515.121640(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > [...] > > >> 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. > > > > Non sequitur: > > > > Logical Implication > > p q p -> q > > F F T *1) > > F T T > > T F F *2) > > T T T > > > I did not prove the truth of the implication (which is true from the > definition of the diagonal in the EIT and from the linear ordering of > the finite elements of N). If the diagonal is defined as containing every element that is in any line, that does not require that there be any line containing all those elements. As a model in ZFC or NBG, take any complete path in an infinite binary tree as the diagonal, and take the finite subpath from the root node to any other node of that path as a line. Then one has a model within ZFC or NBG in which all the parts of WM's EIT are represented properly, and one in which every line has a finite number of edges but the diagonal full path has infintiely many edges. > I proved that the premise is false. Take a > bit time to understand it. If not, take a bit longer. If not at all, > try to study something other than math. Above I take something that WM has by implication allowed, an infinite binary tree, and show that it contains a model of WM's EIT which proves that there are only finite lines but an infinite diagonal, and also that there are infinitely many finite naturals in a system that WM has accepted.
From: Virgil on 19 Dec 2006 23:27 In article <1166545363.371163.143890(a)i12g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > Albrecht wrote: > > > > > > > > but in the same time |N isn't complete > > > > > since |N is bijectable to the diagonal and the diagonal is never > > > > > complete since there is no complete line (number) which covers the > > > > > diagonal. > > > > > > > > No, the claim is that there is no line that covers the diagonal. > > > > > > Which results in the conclusion that there is no diagonal. > > > > 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. WM again loses track of the point at issue. No one wants anything but finite numbers IN the diagonal, but we all want infinitely many of them. And since WM has used infinite binary trees as justifying his arguments, he must allow that a complete path in such a tree is like a diagonal of WM's EIT with its edges being the last members of the lines of that EIT. Which gives every line being finite but the diagonal being infinite. Every edge is in some line but no line contains all edges.
From: Virgil on 19 Dec 2006 23:36
In article <1166545419.868255.124160(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > >> mueckenh(a)rz.fh-augsburg.de wrote: > > >> > William Hughes schrieb: > > >> [...] > > >> >> 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. [(*)] > > > 6. You shall refocus on a proof of (*). > > Every element of the diagonal is n some line. Do you agree? In an infinite binary tree, every edge of an infinite path is n edges from the root. Do you agree? > Every element of the diagonal is a finite natural number, by > definition. Every edge is a finite natural umber of edges from the root. Do you agree? > > 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. There are only finite edges in the path, but infinitely many of them. Do you agree? > > If you cannot understand this proof, then simply try harder. Precisely my advice to WM. > If you > don't succeed but nevertheless are convinced of the contrary, remember: > N is a well ordered set. So there must be a least number n of the > subset for which my statement is not true. Try to find it. Since we started with infinite paths, to which WM has agreed in many posts, and have shown that every edge is only a finite distance, in edge lengths, from the root, WM's arguments that there are only finitely many naturals to count those edges with must fail. |