Cantor Confusion
in [Math]
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From: WM on 4 Jun 2007 14:44 On 4 Jun., 01:34, Virgil <vir...(a)comcast.net> wrote: > In article <1180903562.459979.290...(a)q69g2000hsb.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 26 Mai, 04:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1179931587.894431.146...(a)u30g2000hsc.googlegroups.com> WM > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > On 22 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > In article <1179747854.168331.52...(a)x35g2000prf.googlegroups.com> WM > > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > > > On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > In article > > > > > > > <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM > > > > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > > > > > Its is more. You cannot answer the question whether the > > > > > > > > numbers P = > > > > > > > > [pi*10^10^100] and P' = P with the last digit replaced by 3 > > > > > > > > nsatisfy P < P'. > > > > > > > > Yes, so what? Your distinction is just terminology, and not more > > > > > > > than > > > > > > > that. > > > > > > > It is by far more than a difference in terminology if one will > > > > > > never > > > > > > or always be able to answer a question. > > > > > > Makes absolutely no sense. You have some implicit notion of number in > > > > > mind > > > > > that I do not have. > > > > > Trichotomy. > > > > But there is trichotomy. We know that either P < P' or that P = P' or that > > > P > P'. But we do not have the means to determine which of the three > > > holds. > > > We know nothing about that number. > > Speak only for yourself, WM. If you know more than me about this number, please don't hesitate to publish it. > > ... > > > > Each node splits a set of paths in two sets of paths. When you > > > come in at a node with a set of K paths, you come out with two sets of > > > K/2 paths (this is a bit informal). If K is infinite, so is K/2. And > > > if K is uncountable, so is K/2. So all this splitting does show nothing. > > > And it is not the case that each node separates one more path. If that > > > were the case you should be able to indicate what path is separated by > > > the root node. > > > Every path which can be identified must be separated from all other > > paths. For this sake there must be as many nodes as separated paths. > > That would be the case provided for each path there was a SINGLE node > separating it from all other paths. No. It is true because for every other path q which you copare with p there must be at least a single node. Brefly: Every separated path comes into being only by a noded Therefore there cannot be more separated paths than nodes. Regards, WM
From: Virgil on 4 Jun 2007 14:49 In article <1180966798.913696.131020(a)n4g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 4 Jun., 03:30, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1180903562.459979.290...(a)q69g2000hsb.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > > On 26 Mai, 04:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > Trichotomy. > > > > > > > > But there is trichotomy. We know that either P < P' or that P = P' or > > > > that > > > > P > P'. But we do not have the means to determine which of the three > > > > holds. > > > > > > We know nothing about that number. > > > > Oh. Above I wrote what we know about it. In mathematics both P and P' > > are natural numbers. > > Mathematicians seem to be very short sighted. If they see a great many > of numbers, they quickly assume that here are infinitely many. It is > simply a ridiculous claim. There are not more than 10^100 numbers, and > there are many natural numbers less that 10^10^100 which do not exist > and by no means can be brought to existence. There is a critical difference between the notions of physical existence and that of mathematical existence. Physically, there is no such thing as a triangle, or a line, or a point, but mathematically all of these "exist". So that these, like WM, who would limit mathematical existence by the limits of physical existence, are mathematically clueless. Mathematics is inevitably about ideals that need not and usually cannot be matched physically. To object to that is to misunderstand mathematics. > I need not a bijection in order to show that a set is countable. You need to show that there exists a surjection from the naturals to the set in question, or an injection in the reverse direction, , as one or the other is required by the definition. > Remember the set of constructible numbers. There is no bijection with > N - there is no list. I need merely the simple fact that there can be > no more separated paths than separations in the tree. This idiot notion of "separations", at least as pushed by WM, makes a number of false assumptions about what separates what in a CIBT. The truth is, in any CIBT: Any finite set of nodes, if it acheives any sort of separation at all, only separates one uncountable set of paths from another uncountable set of paths, each set being easily bijectable with the set of all nodes. It takes an infinite set of nodes in any one path to separate it from all other paths. > > > > > > No. Each node splits a set of paths in two sets of paths. When you > > > > come in at a node with a set of K paths, you come out with two sets of > > > > K/2 paths (this is a bit informal). If K is infinite, so is K/2. And > > > > if K is uncountable, so is K/2. So all this splitting does show > > > > nothing. > > > > And it is not the case that each node separates one more path. If > > > > that > > > > were the case you should be able to indicate what path is separated by > > > > the root node. > > > > > > Every path which can be identified must be separated from all other > > > paths. For this sake there must be as many nodes as separated paths. > > > > The first is true. For the second you provide no proof, because it > > assumes that for each path there is a particular node where it separates > > from all other paths. > > Wrong. For every path there is a node where it separates from all > paths considered up to a certain level. But "all paths up to a certain level" never includes all other paths unless there is a LAST level, which does not occur in CIBTs. > > > That is false. > > That means that there must be as many nodes as paths for every finite > segment of the tree, but not for the whole tree? You consider ghost > paths, which do not exist. In WM's world, there may be ghost trees, but in ZF and NBG, there are actual CIBTs which have properties that WM is apparently not capable of analysing. > > > The truth is that for each two > > paths there is a particular node where they separate. > > It is useless to follow this fancy discussion. Fact is that there can > be no more separated paths than nodes. That "truth" if false in ZF and NBG. > Separations happen by nodes. > Every node generates exactly one separation. But what is separated by such a one node separation is the set of paths branching left at the node from the set of paths branching right from that node, and each of these is equinumerous with the set of all paths of the entire CIBT. WM is still trying to argue that properties dependent on the finiteness of a tree must also be properties of infinite trees. > > Every node is the end of a finite path. Except that in CIBTs there are no finite paths at all. There are paths of finite subtrees, but they are not paths of the CIBT itself. > There is simply nothing left > to prove the existence of the other (even uncountably many) paths. In finite trees, of course not, but finite trees are not infinite. > Therefore they are ghost paths. In mathematics, what are ghosts to WM can be quite as real as anything else. It is WM who is out of step with mathematical reality.
From: Virgil on 4 Jun 2007 15:49 In article <1180981322.729241.290610(a)k79g2000hse.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 4 Jun., 01:36, "RLG" <J...(a)Goldofo.com> wrote: > > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > > > > > Are > > you also trying to argue that Cantor's theorem is false? > > It is false. Only in WM's NathUnRealism. > Consider the problem with double representation of some > rationals. This is standardly, and quite easily, dealt with by a construction rle for the diagonal so as not to allow it to equal any number having a dual representation, whether listed or not. Long and irrelevant argument (based on the delusion that dual representation matters) snippped. > I cannot understand, why the general problem with LIM{n-->oo} (a_n - > b_n) * 10^-n = 0 has not been discovered yet. It has long since been discovered to be no problem. At least as far as the Cantor proof is concerned. Why WM keeps deluding himself that it is a problem is his problem, not ours.
From: William Hughes on 4 Jun 2007 16:01 On Jun 4, 2:22 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > The height of the Cantor list is said to be actually > infinite, the width, however, is not. Only in Wolkenmuekenheim. The height and the width are both suprema and are both infinite. The difference is that the set of column heights attains its supremum, the set of line widths does not. Since the bijection does not involve the supremum, this is not a problem. - William Hughes
From: Virgil on 4 Jun 2007 16:02
In article <1180981322.729241.290610(a)k79g2000hse.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 4 Jun., 01:36, "RLG" <J...(a)Goldofo.com> wrote: > > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > > > If I have understood you correctly, you seem to be suggesting that the > > naturals should be considered as a proper class instead of a set. This > > seems a bit unnatural to me but why are you emphasizing this diagonal? > > It cannot be actually infinite unless there is also an actually > infinite line. That is the same false argument as that there cannot be infinitely many naturals unless there is an infinite natural. The height of the Cantor list is said to be actually > infinite, the width, however, is not. That may be what WM says, but it is the same false argument as as that there cannot be infinitely many naturals unless there is an infinite natural. > By the diagonal height and width > are bijected. Therefore both are actually infinite or both are not. It > is simple to see. The "width", being no more than the set of all widths of the separate finite lines, is the infinite set of naturals. > > > Are > > you also trying to argue that Cantor's theorem is false? > > It is false. Consider the problem with double representation of some > rationals. That was taken care of well before WM was as much as a gleam in his father's eye. Cantor's original "diagonal" argument was about infinite binary strings, for which there is no such thing as "dual representation". And until WM has successfully faulted that proof, the decimal proof easily stands as equally valid. But WM cannot fault that proof, and if WM understood the decimal proof, he would see that his futile attempts to fault it are equally fallacious. |