Cantor Confusion
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From: Virgil on 10 Jan 2007 15:28 In article <1168444428.912929.42030(a)77g2000hsv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Not in ZFC or NBG, though the "diagonal" does in both. > Does the tallest man exist? That is, at best, highly time dependent. > > > > > > > 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. In ZFC and NBG it is obviously false. > > > > If you assume actual infinity then L_D > > does not exist. > > It does exist, in potential infinity. But it is not fixed. Then it is not a set according to any standard definition of sets. If WM wants to invent a different name for what he has been misnaming sets, he is free to do so, but in mathematics, sets are not the vaguenesses WM claims. > > In actual infinity (everything including L_D being fixed) it is not > possile to find L_D. And that's the way it is in mathematics.
From: Virgil on 10 Jan 2007 15:46 In article <1168445349.094021.127750(a)p59g2000hsd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. It contradicts WM's assumptions, but that is no problem as there has been no axiom system presented in which his assumptions hold. > > 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) ... > Not necessarily. The set of nodes of one binary tree many be represented by { A; AL,AR; ALL,ALR, ARL, ARR; ...} and those of another by { B; BL,BR; BLL,BLR, BRL, BRR; ...} so their intersection is empty. In which case, the union of the trees will not even be a tree. WM is talking about something like a direct limit of trees (with morphisms being injections preserving root and branching tree structure) not a union of them. http://en.wikipedia.org/wiki/Direct_limit#Formal_definition
From: Franziska Neugebauer on 10 Jan 2007 15:57 Virgil wrote: > WM is talking about something like a direct limit of trees (with > morphisms being injections preserving root and branching tree > structure) not a union of them. > > http://en.wikipedia.org/wiki/Direct_limit#Formal_definition Apropos wikipedia. WM is desparately trying to infiltrate wikipedia: http://de.wikipedia.org/wiki/Spezial:Beitr%C3%A4ge/W._Mueckenheim Fortunately some editors sussed out that he his lingo is not established terminology. F. N. -- xyz
From: Dik T. Winter on 10 Jan 2007 22:19 In article <1168444071.169167.287510(a)i39g2000hsf.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. May be. > 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. That still does *not* show that the union of the sets of paths in the finite trees contains infinite paths. You still fail to provide a proper definition of the union of trees. So I can only guess. Let us assume two finite trees, where I number the nodes: 1 1 / \ / \ / \ / \ / \ / \ 2 3 2 3 / \ / \ / \ / \ 4 5 6 7 / \ / \ / \ / \ We can easily identify like numbered nodes from both trees with each other. And so, with respect to sets of nodes, the union of the two is the tree on the right hand side. With a bit of imagination we can identify the edge going to the left from 2 in the left-hand tree with the edge going from 2 to 4 on the right-hand tree, and in that way see that also with respect to edges, the union of the sets of edges is the set of edges of the tree on the right hand side. If we accept finite paths (and assume that they all start at node 1), there are 6 paths in the tree on the left and 14 paths in the tree on the right. And in this way we accept that the union of the sets of paths is the set of paths on the right. Do you agree with it so far? If you agree, there is however one problem with the complete infinite tree. *None* of the constituent threes contains an infinite path, and so the union of the sets of paths also does not contain an infinite path. > > 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. I repeat my statement. > > > 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. Makes no sense. Elaborate, please. > 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. You are wrong. The union of sets of paths is a set of paths were each of the paths is element of at least one of the sets that define the union. If there is a constituent of the union that contains a particular path, that path is element of the union set. If there is none, that path is not an element of the union. When taking the union of sets of paths there is no consideration needed about edges. > > > 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 that is just what you can find when you allow the axiom of infinity. Suppose the set X consisting of all the finite natural numbers plus a single infinite number. That set does not satisfy the Peano axioms, as a set with a largest element *can not* satisfy the Peano axioms. Simply because there is no successor of the largest element. Now remove that largest element. If there is still a largest element, the Peano axioms can not be satisfied. However, the axiom of infinity states that there is at least one set that satisfies the Peano axioms, and that N is the smallest of those sets. You are claiming that there is no such set. Yes, I know that you are thinking in terms of growing sets, but those are *not* part of set theory. You can start your own branch of set theory with growing sets, but, please, call them something else to avoid confusion. And do not think that using them you can show contradictions in set theory. > 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,...}. Elaborate on the difference. If I use the von Neumann definition, I see no difference, it is only a switch of base (0 or 1), or of labelling. And switching base in the von Neumann definition is quite trivial. > 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. Sorry, the discussion was about the union of *sets* of paths. > 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. I think I was talking about sets of paths. > 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. I wonder. It makes no sense to me. > > 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. Eh? So the union of 0.1 and 0.11 is a set of two finite paths, 0.1 and 0.11? > > 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. I see a line with marbles, beneath it another line of marbles and below that a third line of marbles. What is the meaning? I may add that within the community, I referred to, the question of quantity was irrelevant. Within that culture, quantity was too abstract to comprehend. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: imaginatorium on 10 Jan 2007 23:40
Andy Smith wrote: > Or more easily just extend Peano to the negative numbers. > > Then I think you can show that -1 = infinity > -2 = infinity -1 > etc. > > (which solves my problem of a label for the last element > in an infinite sequence). Hmm. Of course, you could define anything you like, using any names you like, as long as other people can follow your writing. (If you want to do sollipsistic mathematics, no need to bother with Usenet) But I wonder why you use the name "infinite sequence" here? I would think of this expression - more or less by definition - as meaning "a sequence with no end, no last element". In this case there is hardly a need for a label for something that doesn't exist. Brian Chandler http://imaginatorium.org |