Cantor Confusion
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From: mueckenh on 19 Dec 2006 06:12 Math1723 schrieb: > Han de Bruijn wrote: > > > True, but you are ignoring Virgi's adjective "sound". Calculus existed > > > way back during the time of Newton and Leibniz, but you could hardly > > > call their use of the infinite and infinitessimals at all "sound" by > > > today's standards. It wasn't until Bolzano and Weierstrass made things > > > truly rigorous in the 19th century was Calculus anywhere near sound. > > > > Allright. And they should have _stopped_ at this point in time. > > I agree. And what's with these people saying the Earth is round? > Geez, they are going too far. The Earth is round. That can be proved and has been proved - contrary to the consistency of set theory. Regards, WM
From: mueckenh on 19 Dec 2006 06:14 Virgil schrieb: > > > Note that you have already given that there are as many paths in the > > > tree as there are real numbers in [0,1]. We are investigating how we > > > can construct a surjection of the naturals onto the reals. > > > > But we do not presuppose that the reals are uncountable. At least we do > > not presuppose that ZFC is free of contradictions which may result in > > two valid proofs with different results. > > But WM presupposes things that do contradict ZFC. I presuppose the existence of irrational numbers as infinite paths in the tree which differ from other paths, i.e., which exist as individual entities. I conclude that, if their cardinality is to be 2^aleph0, then they cannot exist and their cardinality is zero. > > > > > > Your edges are not > > > being broken into 2 "shares" each; they are being divided into an > > > infinite number of shares, and that is very different from 2 shares! > > > > For this sake we have mathematics, I mean really correct mathematics, > > not based on dubious infinite sets, which states that 1 + 1/2 + 1/4 + > > ... = 2. > > That is not an equality but a hidden limit statement: > Lim_{n --> oo} Sum[k=0..n: 1/2^k] = 2 Set theory veils even the most simple facts. We need not a limit in order to calculate 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n. And that is sufficient. Regards, WM
From: mueckenh on 19 Dec 2006 06:15 Dik T. Winter schrieb: > In article <1166452045.777425.20310(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > >> > > > All elements that can be shown to exist in the diagonal can be > > > >> > > > shown to exist in one single line. [(P1)] > > > > > > This proposition P1 has _not_ yet been proved (shown). > ... > > > Upto now we have no line L_D at all since P1 has not been proved. > > > > > There is nothing to prove in a definition. I remember you said a > > definition cannot be wrong. So it cannot be proved. > > P1 is not a definition. It is the definition that, in the Equilateral Infinite Triangle, the diagonal consists of the ends of lines a_nn. A line has elements a_kn with k =< n and n eps N --- with ALL n eps N. Regards,WM
From: mueckenh on 19 Dec 2006 06:18 Dik T. Winter schrieb: > In article <1166453622.108599.302270(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > And what if m is infinite? How do you divide? How do you recombine? > > > Suppose I have an edge that is divided in countably many shares. Now > > > suppose I select the even numbered shares and recombine them. Do I > > > now have an edge? And what when I do the same with the odd numbered > > > shares, do I now also have an edge? If that is not the case, why not? > > > And if that is the case, please explain. > > > > We need not divide anything by infinite numbers, because there are only > > finite numbers n in the sum > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... > > Every level of the tree has a finite number n. For finite n also 2^n is > > finite. > > I do not understand the connection. Through every edge there go infinitely > many paths. So to give each path a share of the edge we have to divide the > edge in infinitely many shares. We will find out what is infinite. In any case each level has a finite number and the sum of edges inherited by one complete path and by no other path is given by: 1 + 1/2 + 1/4 + ...+ 1/2^n + ... = 2. For the evaluation of this *sum* we need *no limit* but only the computation 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > I would not like to call such a recombination of shares an edge, because > > > it is *not* an edge in your original tree. I would at most call it a > > > collection of shares. And it is quite trivial to split an edge in an > > > infinite number of shares and to recombine those shares in an infinite > > > number of collections of an infinite number of shares. > > > > There are no infinite numbers in 1 + 1/2 + 1/4 + ... > > But each edge is divided in infinitely many shares. It is irrelevant what is infinity here. The sum of edges inherited by one complete path and by no other path is given by 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n. > > > Of course. There is only the finite case for natural numbers. There are > > no infinite numbers in N. > > I do not state so. But through each edge go infinitely many paths. > May be we will find out later on how many paths go through each edge. First let's try to count 1 + 1/2 + 1/4 + ...+ 1/2^n which is the same as to calculate 2 - 1/2^n. > > > > there exists an Hamel basis for the reals as a vector space over > > > Q. And using that basis it is reasonably simple to well order the reals. > > > > Why does such an ordering definitely *not* exist? No book in any > > library of the world (incuding all extraterrestrical cultures) contains > > it, neither a formula, nor a recursion and, obviously, not a list. > > Obviously you need AC to make it existing. So only in that version of > mathematics were AC holds does it exist. So let it hold and make it existing. > > > > In geometry the > > > parallel axiom tells us there is only single line. Abandoning that > > > axiom gives us different kinds of geometry. It is similar with AC. > > > If we accept it we get a different kind of set theory (where every set > > > can be well-ordered) from a version where we do not accept AC. > > > > > > Obviously with AC it has been shown that there is a Hamel basis and that > > > immediately leads to a well-ordering. > > > > Could you post it, please? > > Given the Hamel basis, do some lexicographic ordering of the reals as a > vector space over the rationals. > I cannot do it (and it has been proved that nobody can do it because there is no *definable* well-ordering). So I would prefer you demonstrate it. Regards, WM
From: Franziska Neugebauer on 19 Dec 2006 06:21
mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: >> In article <1166452045.777425.20310(a)79g2000cws.googlegroups.com> >> mueckenh(a)rz.fh-augsburg.de writes: ... >> > > >> > > > All elements that can be shown to exist in the diagonal >> > > >> > > > can be >> > > >> > > > shown to exist in one single line. [(P1)] >> > > >> > > This proposition P1 has _not_ yet been proved (shown). >> ... >> > > Upto now we have no line L_D at all since P1 has not been >> > > proved. >> > > >> > There is nothing to prove in a definition. I remember you said a >> > definition cannot be wrong. So it cannot be proved. >> >> P1 is not a definition. > > It is the definition that, in the Equilateral Infinite Triangle, the > diagonal consists of the ends of lines a_nn. A line has elements a_kn > with k =< n and n eps N --- with ALL n eps N. Please learn what the term "definition" means. F. N. -- xyz |