Cantor Confusion
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From: Dave Seaman on 26 Jan 2007 08:21 On Fri, 26 Jan 2007 13:02:29 GMT, Andy Smith wrote: > Dave Seaman <dseaman(a)no.such.host> writes >>> So if we can show that there is no one to one mapping between N and R >>> taken in some given order, then that would imply that there is no >>> bijection R->N ? > I should have said/ meant to say "So if we can show that there is no > one to one and onto mapping between N and the elements of R ..." >>But there is a quite natural one-to-one mapping (injection) N->R. The >>point is that there is no injection going the other way, and from this we >>conclude that there is no bijection. > This is still following on from trying to show that the reals are > uncountable based on trying to count them systematically, showing that > that fails in that instance, and trying to infer that there is no > injection R->N. (because, for me at any rate, that is a direct approach, > and would provide an explanation meaningful to me as to why the reals > are uncountable). Showing that one particular mapping fails to be a bijection is not sufficient. It doesn't rule out the possibility that some other mapping might be a bijection. The point of the diagonal argument is that addresses all possible mappings f: N->R at once, by showing that each one fails to be a bijection. > even if one buys the argument above, there is still an issue as to > whether the failure of the counting scheme might not be circumvented by > some other means i.e. does the failure of the counting scheme show that > the reals taken in that order are truly uncountable. Uncountability has nothing to do with order. > I would say yes, > because there is no scope for any more compact non-infinite > representation of the reals than their infinite binary expansions. Can you prove that? -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: mueckenh on 26 Jan 2007 09:35 On 26 Jan., 13:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1169804998.026115.299...(a)q2g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > On 25 Jan., 17:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > > > and S is the set of corresponding sequences > > > > > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > > > > > then L cannot be larger than S, because there cannot be more > > > > > > > limits than sequences. > ... > > > Yes? The elements of S as written above are the a_ik for i and k natural > > > numbers. > > > > No. An element of a set S is written as x in S = {x| ...}. > > Yes, if there is a single x. I defined a sequence as an element of the "set of corresponding sequences". Therefore there are single elements, as usual, written in front of the "|". The k-th element is a_1k, a_2k, a_3k, ... >You can also write: > S = { x_i, y_i | i in N } > meaning S contains as elements all x_i and y_i for i in N. Your notation > is a novelty invented by you. New ideas often require new notations. > > > > So the elements are elements from {0, 1}. S *does* contain > > > those paths, but not as elements, but as subsets. S is not the set of > > > "corresponding sequences". > > > > The set of all cars contains every car as an element. But from the > > elements of this set we can form the set of all brands. The set of all > > brands contains 10 or 20 elements only. > > Yes, what is the relevance? It was an example to explain my notation. Now it is no longer necessary. > > > > You specifically wrote: > > > > However, if L is a set of > > > > limits L_k > > > > L = { L_k | k in N } > > > > and S is the set of corresponding sequences > > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > > then L cannot be larger than S, because there cannot be more limits > > > > than sequences. > > > this makes no sense if S is *not* a set of sequences. > > > > S is a set of sequences. > > You say so, inventing completely new notation. Sets of sequences are not so new. > > > > Consider the > > > following: > > > L = { L_k | k in {0, 1, 2}} > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in {0, 1, 2} } > > > If S is a set the cardinality of S is at most 2. > > > > Cardinality is |S| = 3. Its elements are the sequences number k = 0, 1 > > , and 2. > > Wrong. Oh, I thought you had understood? A set of three sequences has cardinality 3. > > > > If S is an ordered > > > *multi-set*, the ordinality is omega. The cardinality of L is 3. > > > S is certainly not an ordered set because it contains multiple identical > > > elements. > > > > S is ordered by the numbers k. > > Wrong. So the sequences S_0, S_1, S_2 are unordered? Regards, WM
From: mueckenh on 26 Jan 2007 09:44 On 26 Jan., 13:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1169806783.611087.268...(a)a75g2000cwd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > On 26 Jan., 02:41, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > In the von Neumann model. But that is a model about ordinal numbers, that > > > start at 0. There are a few subtle differences. I think you are a bit > > > confused here. In the von Neumann model we have that an ordinal number > > > 'k' is the set of all its predecessors ( {0, 1, 2, ..., k-1), if k is not > > > a limit ordinal). And so the union of two ordinals is the larger one. > > > In this case, the ordinal 'k', is also the ordinal number of the set of > > > predecessors. When you shift to '1' base, the latter statement is no > > > longer true. > > > > It has nothing to do with sophisticated models at all. They only have > > to simulate the basic fact. A natural number in its most basic > > representation, namely unary or unadic, simply is the superset of all > > smaller numbers and a subset of all larger numbers. > > In that case you have first to *define* what you mean with "union of > representations". There is no definition for it in mathematics. For natural numbers c is the same as <. 2 c 3 is the same as 2 < 3. This is proven by II + I = III. Perhaps some present mathematicians do not remember these things, because AC and some inaccessible ordinals occupy their brains too much. > > > That is the origin > > which has to be simulated by every correct theory of natural numbers. > > II is a subset of III. > > By what definition? In what way are II and III sets? In that way which need no further declaratiojn. Put three nuts in your hat. Then you can see the set. > > > This kowledge as to be taken into account when > > denoting these numbers by 2 and 3. It has been done by the notation 2 < > > 3. > > In the von Neumann model, indeed. a < b if a subset b. That is the > definition of <. In the Peano axioms < can also be defined, using the > successor function, but there it has not necessarily anything to do with > subsetting. In unary representation it has. In other representations it has to be defined. Regards, WM
From: mueckenh on 26 Jan 2007 09:50 On 26 Jan., 11:11, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > On 25 Jan., 15:57, Franziska Neugebauer > > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > >> > If T(oo) as defined by me is T as defined by me, concerning the set > >> > of nodes and edges, > > >>There is no room in equality for "concerning" this or that. Either two > >> symbols refer to the same entity or they do not. > > > That is not so obvious. That *is* obvious. Indeed? Every function denoted by f is the same one? Even every pair of functions with f(x_1) = g(x_1) is identical? However, concerning trees, Virgil is obviously wrong. I think so too. Regards, WM
From: mueckenh on 26 Jan 2007 10:03
Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > On 25 Jan., 15:34, Franziska Neugebauer > > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > context: > >> ,----[ <45b5ec2c$0$97243$892e7...(a)authen.yellow.readfreenews.net> ] > >> | >> Again: Your notations > >> | >> > >> | >> T(1) U T(2) U ... > >> | >> > >> | >> and > >> | >> > >> | >> U {T(i) | i e N } > >> | >> > >> | >> are undefined. > >> | > > >> | > You are in error. The union of the trees T(n) and T(n+1) is > >> | > defined. n is a natural number. Therefore the union of all finite > >> | > trees is defined. > >> | > >> | You have misunderstood the induction principle. It is not made for > >> | "counting over to the infinite". > >> `---- Look: Above, there is written "i in N", not "i = N". The latter would be missed by induction. > > You have obviously been interested in the "union of all finite trees" > which still waits for a definition. I defined level n and defined the bijection of levels and natural numbers. That is enough. > > > It is irrelevant for the consideraton of trees. > > It is highly relevant to show where your reasoning is malfunctioning. No. You only try to huddle around, avoiding any concrete discussion. You cannot make me believe that your intelligence is insufficient to understand this discussion. > > > In particular we can prove by induction that the tree, formally called > > T(oo), > > 1. Define T(oo). Has been done. > 2. Prove that the "formally called" entity T(oo) is really a tree. T(oo) is a tree by definition. > > > contains all paths if it contains all levels enumerated by > > natural numbers. > > 1. Define "path". Look at my tree. > 2. Define "contains". Look at Wiki: subset. > > > We have no use for your set N in the tree (in case it > > should differ significantly from the ensmble of all natural numbers.) > > Define "ens[e]mble of all natural numbers". > With pleasure (though it is not simple). The "ensemble of all natural numbers" is a somehing which cannot be named without raising wrong impressions. We can only describe it by its effect. If you have this entity, then you have every natural number. If you loose this entity, then you loose nothing but natural numbers. Regards, WM |