Cantor Confusion
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From: Virgil on 18 Dec 2006 15:11 In article <1166451538.861033.303300(a)48g2000cwx.googlegroups.com>, 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. That is only the case in finite trees. It fails miserably in infinite binary trees in which no path has a last node or last edge.
From: Virgil on 18 Dec 2006 15:22 In article <1166452045.777425.20310(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > >> > > > 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). > > This proposition is the definition of the diagonal. Propositions do not define things. The definition of each line is as the union of all previous lines plus one additional member. The definition of the diagonal is the union of all lines endlessly many lines. Because no line can contain the next line, but the diagonal contains them all, no line can contain the diagonal. > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > contained in the diagonal. > > > > The Diagonal is unbounded thus any _assumed_ L_D is not bounded, too. > > Correct is: > If the diagonal is unbounded then any _assumed_ L_D is not bounded, > too. But any assumed L_D must be an L_n, for some n, as these are the only lines possible. And every L_n is a proper subset of an L_{n+1}. So no L_D can exist. > > > > > If your L_D does not contain them, then you have the wrong L_D. > > > > 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. But definitions need not be instanciated. E.g. creating a definition of a unicorn does not mean that any unicorns have to exist, so that defining an L_D does not mean that any L_D has to exist. So in addition to defining an L_D, WM must then prove that one must exist before he can derive anything from his assumption that it does exist. Learn a little common sense, WM.
From: Newberry on 18 Dec 2006 15:32 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? > > > > 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? > > Regards, WM
From: Virgil on 18 Dec 2006 15:35 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. > > > 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. Now there is a REAL self-contradiction. > > > 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. > > He never did tell me how I can divide money among an infinite number of > > persons. > > Perhaps you missed that lesson? > > > But now you are apparently stating that some of the shares can > > be zero. In your tree, what shares are zero and what shares are non-zero? > > It is possible to divide an object into equal shares or into non-equal > shares. Compare a birthday cake and some guests. In principle it is > possible to give nothing to some guest. Then his share does not exist > or, by modern terms, is empty. But all the shares of an edge in my tree > are divided in equal parts, because of justice and because of ease of > calculation. 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? Even in ZFC and NBG, we know better than that.
From: Virgil on 18 Dec 2006 15:43
In article <1166454974.965741.189880(a)80g2000cwy.googlegroups.com>, 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. That is not so obvious that it does not require proof. And any attempt at proof will reveal that WM is relying on secret assumptions which are false in ZFC and NBG and many other set theories. > >... Therefore L_D does not exist. > > I think you can agree to the statement: > The diagonal does, by definition, not contain any element which is > missing in every line. > > Now assume that at least two lines were required to contain all the > elements of the diagonal. We assume much more, that infinitely many lines are needed. > All the lines are finite, so one of them must > be smaller than the other. You have assumed /at least 2/ so there may be more than 2 and thus need not be a "one or the other" situation. If you assume a finite number of lines then, and only then, is there an L_D = L_max(n in N). But that assumption is what you are trying to prove. So your argument is, as usual, circular. |