Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 19 Dec 2006 06:49 Newberry schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Newberry schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Newberry schrieb: > > > > > > > > > Sorry that I joined a bit late. > > > > Doesn't matter. > > > > > > > > >Are you saying that (in an infinite > > > > > binary tree) the set of paths is uncountable but the set of edges is > > > > > countable? > > > > > > > > The set of edges is obviously countable by, e.g., > > > > > > > > 1, 2, > > > > 3,4,5,6, > > > > 7,... > > > > > > > > As no path can separate from another one without the existence of two > > > > more edges, the number of edges is an upper bound for the number of > > > > paths. > > Is this formally provable in ZFC? I don't know. Up to now nobody has found a way. But also nobody has found a contradiction in the tree. > > > > > > > If it is so simple why do you still need the proof by inheritance of > > > shares? > > > > I don't need it. But there are others who do not see the forest because > > of too many trees blocking the view. > > Going back to the shares ... I can sort of see that each path will > acquire enough shares to build two edges. What is supposed to follow > from that? It follows that the cardinal number of paths is not larger than the cardinal number of edges. The edges have cardinal number aleph_0. A subset of the paths can be shown to be in bijection with the real numbers of the interval [0,1]. Regards, WM
From: mueckenh on 19 Dec 2006 06:55 Virgil schrieb: > In article <1166453622.108599.302270(a)j72g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > In article <1166361677.475397.105840(a)l12g2000cwl.googlegroups.com> > > > mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > > > > > The notion "rational relation" is a further development of the > > > > > > notion > > > > > > "relation". While a relation connects elements of a domain to > > > > > > elements > > > > > > of a range, the rational relation connects shares of the elements > > > > > > of > > > > > > the domain to (shares of the) elements of the range. A simple > > > > > > example: > > > > > > > > > > Again, what are "shares"? > > > > > > > > An edge is considered to be a set of shares. You can divide it into m > > > > equal shares. If you divide n edges into m equal shares each, then you > > > > have m * n shares and you can combine m abitrary shares to have one > > > > full edge. As an example use euro and cent. > > > > > > 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. > > Infinitely many paths pass through every edge of the tree, so if any one > of those edges is to be shared among those paths, it must be divided > into infinitely many pieces. Obviously this has been done already if your statement "infinitely many paths pass through every edge of the tree" is correct. > > > > > decimal point here. Let us denote the two equal shares of edge number 1 > > > > by 1#1 and 1#2. > > > > > > What is bewildering is that you are again using a theorem that would be > > > valid in the finite case as being also valid in the infinite case without > > > proof. > > > > > > > Consider you have two coins of 50 cent value each. Then you can buy > > > > something which costs 1 euro. > > > > > > Again, the finite case. > > > > Of course. There is only the finite case for natural numbers. There are > > no infinite numbers in N. > > Then there should be no infinite binary tree in the first place, and all > of WM's claims about such trees are based on his simultaneous allowing > of infinite paths and denial of anything infinite. You believe that infiniteley many finite numbers exist. Why do you doubt that the paths can share the same property? > > > > > 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? > > Only after you prove that the axiom of choice is actually true without > needing to be assumed. Assumption and truth are two different things. That's my saying. > > WM wants to divide something into a countably infinite number of equal > shares. > > If s is the size of one share then WM wants to have s + s + s ...= 1. > > What is the numerical value of s, WM? > 1/s is 2^n with n a natural number. What is the largest finite natural number? Regards, WM
From: mueckenh on 19 Dec 2006 06:57 Virgil schrieb: > In article <1166455723.089710.246540(a)f1g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Newberry schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > As no path can separate from another one without the existence of two > > > > more edges, the number of edges is an upper bound for the number of > > > > paths. > > > > > > If it is so simple why do you still need the proof by inheritance of > > > shares? > > > > 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. Have you never put sqrt(2) = p/q in order to show that this is not possible? Regards, WM
From: mueckenh on 19 Dec 2006 07:00 cbrown(a)cbrownsystems.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > cbrown(a)cbrownsystems.com schrieb: > > > > > > > > > > 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. > > > > > > > > > > "Really correct mathematics" doesn't simply /state/ that "1 + 1/2 + 1/4 > > > + ... = 2". Mathematics /defines/ what the series of symbols "1 + 1/2 + > > > 1/4 + ... = 2" means, and then /proves/ that it is true from these > > > definitions. > > > > > That is the nonsense en vogue today. > > 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. Regards, WM
From: William Hughes on 19 Dec 2006 07:06
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). The existence of all the elements of the diagonal is neither claimed nor used. The "infiniteness" of the diagonal is neither claimed nor used. The contradiction (a line must be both bounded and unbounded) does not require any assumption that you have not made. - William Hughes |