Cantor Confusion
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From: mueckenh on 10 Jan 2007 10:21 William Hughes schrieb: > > > No. It is clear that given any two natural numbers, or indeed any > > > set of natural numbers you can write down, there exists a single > > > line which contains all the natural numbers you wrote down. But this > > > line > > > *can* depend on which set of natural numbers you write > > > down. > > > > Of course. And if you write down all natural numbers, then you write > > down L_D. > > If the set of all natual numbers is complete, then L_D is complete. > > If the set is potentially infinite, then L_D can change. It depends on > > which set of natural numbers you just have. > > L_D cannot change. L_D is a line. A line cannot > change. No. But the property of being the greatest line can and does change. > You are saying that for every maximum integer, n_m, > that can be shown to exist, a line L_M(n_m) > which contains all integers up to and including > this maximum (i.e. all integers that have been shown to > exist at this point) can be shown to exist. > No one is disuputing this. However, you cannot find > a single line L_D that contains every L_M(n_m) > that can be shown to exist. Of course you cannot find that line. It would be actual infinity. Cantor calls it the Vollendet-unendliche (finished infinite) As soon as the previously non-existing number exists, there exist the diagonal up to that number and there exists the line which is as long as the diagonal. Regards, WM
From: William Hughes on 10 Jan 2007 10:43 mueck...(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > No. It is clear that given any two natural numbers, or indeed any > > > > set of natural numbers you can write down, there exists a single > > > > line which contains all the natural numbers you wrote down. But this > > > > line > > > > *can* depend on which set of natural numbers you write > > > > down. > > > > > > Of course. And if you write down all natural numbers, then you write > > > down L_D. > > > If the set of all natual numbers is complete, then L_D is complete. > > > If the set is potentially infinite, then L_D can change. It depends on > > > which set of natural numbers you just have. > > > > L_D cannot change. L_D is a line. A line cannot > > change. > > No. But the property of being the greatest line can and does change. Irrelevant. The question is whether L_D exists. > > > You are saying that for every maximum integer, n_m, > > that can be shown to exist, a line L_M(n_m) > > which contains all integers up to and including > > this maximum (i.e. all integers that have been shown to > > exist at this point) can be shown to exist. > > No one is disuputing this. However, you cannot find > > a single line L_D that contains every L_M(n_m) > > that can be shown to exist. > > Of course you cannot find that line. Your putative proof that assuming actual infinity leads to a contradiction is: It is possible to find L_D If you assume actual infinity then L_D does not exist. Therefore you cannot assume actual infinity. You now admit that it is not possible to find L_D, only the L_M(n_m) . Yes, if you assume actual infinity then the description of the lines L_M(n_m) changes slightly. However, these lines still exist. Assuming actual infinity does not lead to a contradiction. - William Hughes
From: mueckenh on 10 Jan 2007 10:47 Dik T. Winter schrieb: > > The union of all finite trees (= union of all levels) of the tree > > automatically contains all paths, because there is no path or part of a > > path outside of this union. > > And again, in that case the set of paths in the union is not the union of > the sets of paths. I did show that with a finite graph. So I wonder > why you think it holds for an infinite graph (which a tree is)? We are concerned with real numbers which are represented in the tree just as I defined it. Every paths splits at a node into a left one and a right one. In the unin of trees there are all paths or pats of paths (in the sense defined and also the crippled ones) which exist in this union. Nothing can be added unless you add another edge. > Yes, but that does *not* make the *set* of paths the union of the *sets* > of paths in the finite trees. In the infinite union of all finite trees there are all possible paths, because there is no edge which could be added, but paths consist of edges. > > > > Correct. As the union of all finite paths turning always right is > > 0.111... . > > Pray, define the union of paths. But I am talking about *sets* of paths. The union of finite paths is defined like the union of initial segments of indexes. By a lucky accident, the union of all sets of paths are given by the union of all corresponding edges. That's why I devised the tree. > > > > U{n is natural} {n} = N. Because every k in N is in one of the sets used > > > in the union, namely {k}. {0.111...} is *not* in > > > U[ {0.1}, {0.11}, {0.111}, ...} ] because it is in *none* of the sets > > > used in the union. > > > > But {0.111...} is {{0.1}, {0.11}, {0.111}, ...} (see below). > > Oh. > > > > The union of {0.1}, {0.11}, {0.111}, ..., is *not* {0.111...}, it is: > > > {0.1, 0.11, 0.111, 0.1111, ...}. Remember: unions are defined for *sets*. > > > I do not know any definition of union for numbers as these. > > > > > And what is the set {0.1, 0.11, 0.111, 0.1111, ...}, or, translated to > > digits, the set {1,2,3,...}? Is it not {0.111...} or, translated to > > digits, N (or omega)? > > Perhaps. But what is the relevance? The relevance is that the uninon {1,2,3,...} is *not* an infinite set N with cardinal number aleph_0, as long as there is not an "infinite number" *in* the set. But you do not see this small difference. You claim that the union of all finite numbers or of all finite initial segments {1,2,3,...,n} is the infinite set {1,2,3,...}. Now you see the contradiction, I hope. The union of all finite paths belonging to one infinte path in the tree does *not* yield this infinite path. Otherwise the reals were proved countable. Therefore you must add, in addition to all finite segments of that very infinite path, something which you cannot describe. Something, however, which is certainly not a finite segment of that path. Plainly tranlated to the real numbers: You see that the infinite sequence of digits 0.a_1,a_2,a_3,... with finite indexes does not yet establish the real number. You have to add something else. Alas, there is nothing to add. > How about the set {0.1, 0.10, 0.101, > 0.1010, ...}? How do you define the *union* of paths? In a level three > tree I can encounter the paths 0.111 and 0.101. What is their union? Their union is a set of two finite paths 0.111 and 0.101. > > So consider the circles below. Their meaning should be unique. > > For you, apparently. > > > > > > > They will understand, at least by experiment, > > > > > > > > > > > > oo > > > > > > ooo > > > > > > _____ > > > > > > ooooo > > > > > > > > > > You apparently have not done the experiment. No, they will not > > > > > understand, and such experiments have been done. > > > > > > > > Depends on their intelligence, which I don't know. > > > > > > It has nothing to do with intelligence. In their culture they do not use > > > *any* abstraction. They do everything with concrete things only. > > > > In above circles there is no abstraction. > > You think so. I see (in my non-abstracted view) a line with circles, beneath > it another line of circles, a dash and a third line of circles. I have no > idea what the meaning of that is. A picture, apparently. Looks nice. But > what is the meaning? Take marbles or sticks if circles are too abstract. Regards, WM
From: mueckenh on 10 Jan 2007 10:53 William Hughes schrieb: > mueck...(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > No. It is clear that given any two natural numbers, or indeed any > > > > > set of natural numbers you can write down, there exists a single > > > > > line which contains all the natural numbers you wrote down. But this > > > > > line > > > > > *can* depend on which set of natural numbers you write > > > > > down. > > > > > > > > Of course. And if you write down all natural numbers, then you write > > > > down L_D. > > > > If the set of all natual numbers is complete, then L_D is complete. > > > > If the set is potentially infinite, then L_D can change. It depends on > > > > which set of natural numbers you just have. > > > > > > L_D cannot change. L_D is a line. A line cannot > > > change. > > > > No. But the property of being the greatest line can and does change. > > Irrelevant. The question is whether L_D exists. It does. Does the tallest man exist? When did it start, when did it cease? > > > > > > You are saying that for every maximum integer, n_m, > > > that can be shown to exist, a line L_M(n_m) > > > which contains all integers up to and including > > > this maximum (i.e. all integers that have been shown to > > > exist at this point) can be shown to exist. > > > No one is disuputing this. However, you cannot find > > > a single line L_D that contains every L_M(n_m) > > > that can be shown to exist. > > > > Of course you cannot find that line. > > Your putative proof that assuming actual > infinity leads to a contradiction is: > > It is possible to find L_D It is not assumed, but it is obvious that for every given set of natural numbers there is one line containing it. > > If you assume actual infinity then L_D > does not exist. It does exist, in potential infinity. But it is not fixed. That is a property of potential infinity, because every set is finite. > > Therefore you cannot assume actual infinity. > > You now admit that it is not possible to find L_D, In actual infinity (everything including L_D being fixed) it is not possile to find L_D. > > Yes, if you assume actual infinity then the description > of the lines L_M(n_m) changes slightly. However, these > lines still exist. Assuming actual infinity does > not lead to a contradiction. It is a contradiction. Regards, WM
From: mueckenh on 10 Jan 2007 11:09
Dik T. Winter schrieb: > In article <1168351276.094403.80470(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > What is a ghost path? > > > > A ghost path is a path which does not exist in a tree which contains > > the paths of all real numbers the bits of which can be indexed by > > natural numbers. > > That is something I do not understand. Please elaborate. See my recent posting. You need to assume that the complete union of levels of the binary tree does not contain all the paths representing real numbers. Therefoer something must be added, in order to get the complete set of paths. But there is nothing to be added - but ghosts. > > > > > The union of all finite trees is also the union of all last levels. > > > > That is the union of all levels. If some path is missing, then it must > > > > stretch over more than all levels indexed by natural numbers. > > > > > > Ah, again about that diagonal. No, you are wrong. > > > > All path the bits of which can be indexed by natural numbers are in the > > union of all finite trees. > > But *not* in the union of the sets of paths in the finite tree. Yes. The union of all finite trees is not an (actally) infinite tree. See my due explanations of the union of all lines of the EIT: 0.1 0.11 0.111 .... As long as there is not a line with infinitely many digits 1 there is no diagonal with infinitely many digits 1. Conclusion: There is no diagonal with infinitely many digits 1. The axiom of infinity leads to a contradiction. > > > You know it. You cannot point to a bit of any number of [0, 1] which is > > missing. But you claim there were further numbers. > > > > Try to complete the union of all finite trees to the complete infinite > > tree. Which level, node or edge would you add? > > Pray, properly define the union of trees. In my opinion a tree consists > of three sets. First the set E of edges, next the set N of nodes and > finally the set P of paths. What is the union of two trees? With the set of nodes and the definition of edges all other sets are defined. > > > > Why was it not a proof? > > > > Because you used crippled plants which are not under discussion in a > > forest of well grown trees. > > My proof was about the set of paths in the union of finite trees not being > the union of the sets of paths in the finite trees. What was wrong about > that proof? Please, once come up with a proper definition of the union > of two trees. Before you come up with such a definition it is impossible > to even entertain a discussion. Definition: The nodes of the tree are denoted by (0,0) (1,0) (1,1) (2,0a) (2,1a) (2,0b) (2,1b) .... (n,0a) (n,1a) ... The union of all trees up to the n-levels tree is {(0,0)} U {(0,0), (1,0), (1,1)} U .. U {(0,0), (1,0), (1,1), ..., (n,0a) (n,1a) ...} End of definition. Example: The union of the one-level tree and the two-levels tree is {(0,0), (1,0), (1,1)} U {(0,0), (1,0), (1,1), (2,0a), (2,1a), (2,0b), (2,1b)} > > > > I am talking about the union of *sets* of paths, not the union of paths. > > > Do you not see the difference? How do you define the union of paths? > > > And how do you define the union of the *sets* of paths that are in the > > > finite trees? > > > > Which edge of any infinite path is missing? Which bit could be added to > > one of the numbers represented? > > What is the relevance? If you can state that there is some infinite path > in the union of the *sets* of paths in finite trees, you should also be > able to point to a finite tree that contains that infinite path. There is no infinite initial segment of N of the from {1,2,3,...,n}. Nevertheless the union is said to be {1,2,3,...}. > There is > no edge missing, it is only that your infinite path is not a path in any > of the finite trees, so it is not in the union of the sets of paths in > finite trees. So, how can an infinite path be described? How can the corresponding real number be described, if not by all the levels n e N available in the union? > > > > Ok, so you do *not* use the union of the sets of paths but something else. > > > > The material from which the paths are constructed. > > Makes no sense. But is fact. Regards, WM |