Cantor Confusion
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From: Andy Smith on 11 Jan 2007 19:55 > > The problem is your wording. Consider the sequence 1, > -1/2, 1/4, -1/8, > ... Then there is no last element to the sequence, > but the set > > {1, -1/2, 1/4, -1/8, ...} > > certainly contains a smallest element and a largest > elment. > > You have to remember that in mathematics, words (like > "sequence") have > precise technical meanings. You need to use the > jargon correctly or > people will correct you! > Yes, sorry speaking too loosely. Of course no problem with an infinite set having a max and a min. But if you can systematically order the set between max and min that puts its elements in correspondence with the natrural numbers, but has a first and a last element (and there exists to greatest last natural number, trivially from Peano axiom). > > On the Pe ano thing, what happens if you remove the > axiom > > that says 0 is not the successor of any natural > number? > > I'm not sure. I would think you'd have to modify the > induction axiom too > to get something sensible. > > > Then you get negative numbers included as natural > numbers, > > other properties unchanged for the positive > numbers. > > What does the e.g. binary representation of -1, -2 > then > > look like ? > > "-1" is the binary representation of -1. "-10" is the > binary > representation of -2. > > > A horror, with ...1111 as -1. Do -inf and > > +inf converge on the same number? > > Are you asking about IEEE arithmetic or are you > asking about how +oo and > -oo are used in analysis? > > > Can one then just regard -1 > > as the point that one counts round to eventually? > Then > > you could have a start and an end with an infinite > number > > of points in between, was how my uneducated thought > > > processes went, and then I wouldn't have to fret > about > > Zeno. > > I don't follow what is bothering you about Zeno. As > for whether we can > make the integers into a loop, the problem with such > constructions is > how to define the arithmetic operations and whether > the resulting > operations will have the properties that we expect or > want them to. > Yes, I was thinking that in the natural number generation from peano, addition is not defined - as you point out numbers are defined in terms of order of succession. So for negative numbers you can't write -1 - it has to be a string with the property that its successor is 0. But this is all garbage anyway.
From: Dave Seaman on 12 Jan 2007 07:49 On Fri, 12 Jan 2007 05:45:37 EST, Andy Smith wrote: > Perhaps I should have asked, can you have a "countably > infinite" ordered set with a first and last member? Sure. Any countable successor ordinal will do. For example, w+1, where w represents omega, the first transfinite ordinal. The members of w+1 are 0, 1, 2, ..., w. Its first (smallest) member is 0, and its last (largest) is w. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Dik T. Winter on 12 Jan 2007 07:58 In article <1168445349.094021.127750(a)p59g2000hsd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > 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. Oh. But depends on what is done. But even if you unite two finite levels, the set of paths in the union is not the union of the sets of paths of the separate levels. > > > 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: And again you *fail* to argue with the union of sets of paths in mind. But that is something that you actually do use when you are talking about the cardinalities of those sets and draw conclusions from that. > 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. That is your (wrong) thinking. > > > 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. Right. And in that case the union of the sets of paths in individual trees is *not* equal to the set of paths of the united 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)} Ok. Now show that the union of the sets of paths is the set of paths of the union. > > > 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,...}. Yes. There is no infinite *set* of paths in the finite trees, nevertheless is the *set* of paths in the infinite tree infinite. On the other hand, there is no infinite natural in the infinite segments and so there is also no infinite number in their union: N. As there is no infinite path in any of the *sets* of paths of the finite trees, so there is also no infinite path in their union. > > 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? There are infinite paths in the union of the trees. There is *no* infinite path in the union of the sets of paths. Those two things are different. > > > > 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. How than can you draw conclusions about the cardinality of the sets of paths in the union from the cardinalities of the sets of paths in the finite trees? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Lee Rudolph on 12 Jan 2007 08:01 Dave Seaman <dseaman(a)no.such.host> writes: >On Fri, 12 Jan 2007 05:45:37 EST, Andy Smith wrote: > >> Perhaps I should have asked, can you have a "countably >> infinite" ordered set with a first and last member? > >Sure. Any countable successor ordinal will do. For example, w+1, where w >represents omega, the first transfinite ordinal. The members of w+1 are > > 0, 1, 2, ..., w. > >Its first (smallest) member is 0, and its last (largest) is w. For a quite different example of what Andy Smith is asking for (maybe not what he meant to ask for), take the rational numbers between 0 and 1 inclusive. Lee Rudolph
From: Dik T. Winter on 12 Jan 2007 08:01
In article <1168533543.556066.104940(a)p59g2000hsd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > 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. > > Why do you see it different in case of the union of all finite trees or > paths? I am not talking about unions of paths. I am talking about unions of *sets* of paths. That is something completely different. > > > 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. > > representing finite and infinite sequences of numerals of real numbers. But the sets of the paths in the finite trees do *not* contain infinite sequences, and so the union of those sets can not contain infinite sequences. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |