Cantor Confusion
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From: Franziska Neugebauer on 25 Jan 2007 09:34 mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Jan., 21:05, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: [...] >> Using Virgil's trees we have the result that the induced std-tree >> (M(P), E(P)) may be the same if different sets of paths P are given. > > Then it is not what I am talking about. So what _are_ you then talking about? > If T(oo) = T concerning the set of nodes and edges, Axiom of Extensionality: Two sets a and b are equal if A c (c e a <-> c e b) There is no room in equality for "concerning" this or that. > then they are identical with respect to paths too. There is no further > parameter to fix. If you do not accept this identity, then further > discussion is not meaningful. Have you scrapped your plan to show a contradiction in contemporary set theory? >> [...] >> >> >> Since the union of V* is the std-union we know that that U V* >> >> contains only "finite" paths. >> >> > Correct.This is correct _only_ for Virgil's trees. For std-trees >> > there the >> std-union is not a tree and for David's tree-union the tree-union is >> not the std-union and the tree-union of all finite trees is the >> binary tree G. >> [indentation correction:] >> > Hence 0.[10] is _not_ in UV*. >> Using Virgil's trees this is correct. >> Using David's tree-union this is not correct. > > I do not intend to talk about this or that tree which may veil the > facts. If you still want to show some contradiction in contemporary set theory you have to commit to an appropriate formalization of what you consider a tree. > In my proof there are two sorts of trees, solely defined by > nodes and edges. Let the other trees grow, exist, contain, or whatever > they like, when and where they can, if they can. We have many sorts of trees. You should commit yourself to one. >> ,----[ <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". >> `---- >> >> >> You have misunderstood the induction principle. It is not made for >> >> "counting over to the infinite". >> >> > It is valid for all existing natural numbers. Counting over to the >> > infinite is nonsense.[indentation corrected] Counting occurs in the >> > finite. > > What you do _is_ "counting over to the infinite" > > By no means! Every n in N is finite. The set of finite trees V* = { T(n) | n e N } is not a finite set. Induction over its members does not prove a property of V*. This is quite the same as in the case of the infinite set N = { n | n e N }. Induction over n for some property p(n) does not prove p(N). Maybe you forgot that. >> when you claim that >> induction proves existence of U V* where in fact is does only prove >> existence of UV(n) A n e N. > > Yes. > >> V* is not a finite set, i.e. a set of the >> form V(n). Hence induction does not "reach" V*. > > Induction reaches and includes every tree which exists. What *exists* in ZFC is defined by the axioms and the rules of inference. If you still want to show a contradiction _in_ ZFC you must adhere to that axioms and rules. In ZFC/graph theory the tree T(oo) = G (the infinite binary tree) exists but is only "reached" when using the non-std-union of David. When using the std-union of Virgil, the _induced_ tree (M(P), E(P)) is "reached" but the Virgil-tree (i.e. the union of all finite _paths_) is not identical to the paths induces by the infinite binary tree T(oo)/G. I must admit this is probably not simple enough at applied university level but nonetheless valid. None of the trees (Virgil/David) are "reached" by induction. Maybe that WH can help you research this issue not assuming the axiom of infinity (in the framwork of so called potential infinity). >> >> If tree-union was the std-union of ZFC then by the axiom of union >> >> we can state that U V* is defined. But that requires a commitment >> >> to Virgil's definition of trees, which is not the standard-meaning >> >> of trees. >> >> > Virgil dreams of different sets of paths in one and the same tree. >> > I would not believe in his ideas. > >> Virgil has states that there are at least two different sets of path >> induce the _same_ std-tree G. This is no dream or believe. This is >> proven fact. > > As I said above: If you conclude from one and the same tree that there > are different paths, There exist different sets of paths (Virgil's trees) which *induce* the same standard-tree (M(P), E(P)). > i.e., that one of the tree contains paths which the other does not > contain, then we should stop. If you want to ride an equivocation, do so. > I am not willing to waste my time with exlicit nonsense. (Set theory > hitherto was only implicit nonsense.) The nonesense is explicitly of Augsburgian origin. F. N. -- xyz
From: mueckenh on 25 Jan 2007 09:45 On 25 Jan., 13:39, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > 1) If we identify path-lengths with natural numbers, then there is no > > infinite path in any of the finite trees, because there is no infinite > > natural number. Therefore, also the union of all finite trees cannot > > contain an infinite path. > Clearly the tha path-length of an infinite path cannot be a > natural number. The path length is the number of nodes of the path. As long as the nodes are enumerated by natural numbers, the length (distance from the root node) is a natural number. > We are left with > > > 2) If we identify path-length with sets of natural numbers, then there > > are infinite paths in the union tree T(oo). > >However, if L is a set of > > limits L_k > > L = { L_k | k in N } > L is the set of finite limits. > L is countable. Note that the set L does not > contain an infinite limit. Every limit is an entity. > > > 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. > Note that S contains only infinite sequences. Every sequence is in the union T(oo). > > S can be and is larger than L. S is not countable. That again is an unjustified and provably wrong assertion. The sequences belong to the countable union of the finite trees. *Everything* in this union is countable. You could see this by yourself. Please spare such nonsense in future. > > > Different limits for one sequence would mean a path > > splitting at a level n beyond any natural number, i.e., n not in N. > > This is not only impossible but would also lead to undefined limits. > > > Therefore, even if union and limit are two different things (due to the > > schism), both things lead to countably many paths in T(oo). > No. The set of paths in T(oo) is the set of sequences S. > This set is not countable. In a union of trees containing only finite paths, there cannot be a finite path. Regards, WM
From: mueckenh on 25 Jan 2007 09:48 Dik T. Winter schrieb: > In article <1169728742.271970.13430(a)v33g2000cwv.googlegroups.com> "William Hughes" <wpihughes(a)hotmail.com> writes: > > On Jan 25, 3:19 am, mueck...(a)rz.fh-augsburg.de 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. > > > > Note that S contains only infinite sequences. > > S is not a set of sequences. S is an ordered multi-set of the digits > 0 and 1. Or, alternatively, S is one of the sets {0, 1}, {0} or {1}. Sorry. S contains for instance the paths 0.111... and 0,010101... the bits (nodes) of which can be indexed by natural numbers. Regards, WM
From: Franziska Neugebauer on 25 Jan 2007 09:57 mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Jan., 21:05, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > >> > Wrong. If A c B or B c A, then A U B is a tree, namely B or A, >> > respectively.There are no two trees A and B (using >> > std-Kuratowski-pairs:) >> >> A := {{M_A}, {M_A, E_A}} >> B := {{M_B}, {M_B, E_B}} >> >> where A c B or B c A. Trees necessarily have cardinality 2. >> Hence your remark is entirely pointless. > > You will not understand or cannot understand what I say. (You are the > only one here.) This would be a remarkable insight of yours if you opted for the second alternative. > The structure of my trees is given. Therefore even the edges are not > necessary, but a guide for the eye. You forgot a guide for your brain. > Hence, the union of trees is the union of ordered sets of nodes. > If you accept this, then we may continue. If you do not accept this, > then leave it as it is. Not without telling your repeatedly that you have not fully specified (commited) to what exactly and formally defines trees and a unions thereof. Until then you write vacuous verbiage. >> >> 1. You are in error about what is defined first in ZFC. >> >> > I did not say what was first. (There is no time in mathematics.) > >>When I mean "first" I do not mean time but dependence. In ZFC '<' >> depends on 'c' not the other way round. > > It is completely irrelevant why this is so in ZFC. What counts is hat > it is so. Just to recall your claim: ,----[ WM ] | The union of two natural numbers is defined to be the larger one. | This is a set theoretic union. `---- So the union of two natural numbers is _not_ defined to be the larger one but the "larger" of two distint naturals (sets) a and b is defined to be a iff b c a or b iff b c a. So the order _is_ relevant to successfully prove you wrong. >> > I said what is fact. It is possible to define the union of two >> > finite numbers by the maximum of both. > >>It is not the aim of any axiomatic theory to rely on what you see. >> Maybe someone can explain the rules of the game to you. > > I am not interested in a game which is as stupid as you report it. I > am interested whether there are irrational numbers. So you are a skeptic? > And I have found a way to show that they do not exist unless there are > mathematical entities which are unique and nt unique simultaneously. As shown you are riding the equivocation. >> So you _have_ scrapped your plan to show a contradiction _in_ >> contemporary set theory? > > 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 to symbols refer to the same entity or they do not. > then they are identical with respect to paths too. > There is no further parameter to fix. If you do not accept this > identity, then further discussion is not meaningful. E-qui-vo-ca-tion. > Then you may maintain ZFC and may say that it is free of > contradictions, You have not shown any yet. > but you must tolerate that a unique structure may > yield different results, i.e. is not unique. I refuse to tolerate that nonsense. > My goal is reached. Daydreaming? F. N. -- xyz
From: Andy Smith on 25 Jan 2007 10:04
> >When I talked about the reals, I will say it again: > >1) Any systematic method for counting the reals is equivalent (you would >probably say isomorphic under permutation). > >2) I proposed to try to count the reals by progressively increasing the >degree of precision of their representation. To n bits precision, there >are 2^n distinct numbers, requiring 2^n indices. If I wish I can >represent a unique set of indices covering this space by e.g. reflecting >the bit pattern of the reals to the first n bits about the binary point. > >3) For all finite n, no problem. The set of indices are finite, and this >basically is just indexing (a subset of) the rationals. However the >reals require an infinite n and cannot be represented with indices with >a finite number of bits. (Aristotle would say that the natural numbers >are potentially infinite, but the reals are actually infinite). > > I might have added that the purpose of the bit reflection was so that the index associated with each real was not a function of the bit offset n i.e. once counted a real number retains its index. 0.00.. is 0, 0.10.. is 1 etc. -- Andy Smith |