Cantor Confusion
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From: G. Frege on 25 Jan 2007 03:43 On Thu, 25 Jan 2007 08:38:08 GMT, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > > [that] doesn't mean that everything I say is rubbish... > Concerning math, it is. Sorry. F. -- E-mail: info<at>simple-line<dot>de
From: mueckenh on 25 Jan 2007 04:04 On 24 Jan., 17:39, step...(a)nomail.com wrote: > Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > > In message <1169646430.885558.7...(a)m58g2000cwm.googlegroups.com>, > >>What do you mean by a "completed set"? Is an unending sequence such as > >>the following a "completed sequence"? > > >>1, 11, 111, 1111, 11111, 111111, 1111111, ... > > >>If "completed" means "having an end", then this is _not_ "completed", > >>because it continues without end. But if "completed" means "complete" - > >>that is, that there is nothing "missing" - then the sequence above > >>includes every two-ended string of 1s, even though there are an > >>unending number of them. Do you disagree? > > Um, well I thought there was a distinction between the sequence > > 1,2,3,4,n,.. and {1,2,3,4,n,..}. All the terms in the sequence are > > finite, but the set of all of them is infinite. >The key distinction between a sequence and a set is that a sequence > is ordered. The key distinction between a sequence and a set is that a set can contain one and the same element only once, the set {1,1,1,2} is the set {1,2}, while a sequence can contain it as often as you wish. The sequence 1,1,1,2 is different from the sequence 1,2. Order is also possible for sets. Regards, WM
From: mueckenh on 25 Jan 2007 04:13 Virgil schrieb: > In article <1169626572.550075.102890(a)l53g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > > > > > You have to choose whether you want to see a tree as > > > > > > a) the standard tree, T = (M, E) > > > b) the "Virgil"-tree. T = (set of paths having certain properties). > > > > > > Which definition do you choose? > > > > I defined the trees which I use by nodes and edges. > > > > 0. > > / \ > > 0 1 > > / \ / \ > > 0 1 0 1 > > | | | | > > 0 1 0 1 > > | | | | > > 0 1 0 1 > > . . . . > > . . . . > > . . . . > > > > > > This is a tree of type weeping willow, T(1), with edges 0 and 1. It is > > easy to see how the general definition of T(n) looks like but it is > > tedious to write it down here. However, who is unable to understand it, > > will be unable to follow my thoughts and should refrain from trying so. > > > > > > The corresponding cut tree T(1) is > > > > 0. > > / \ > > 0 1 > > > Why not > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > ? > That is a complete binary tree of depth 2 within you WW tree. No. The bottom nodes have not two child nodes but only one. > > And your Weeping Willow Trees are essentially the same as my eventually > constant trees, to which you objected. If they are the same as mine, then I will not object to yours, but I think we should concentrate on not too much definitions. > > > > We see that the union of two trees is the larger one. We see further > > that the union is the set theoretic union. And by enumerating the nodes > > > > 0 > > 1 2 > > 3 4 5 6 > > ... > > > > we prove that the union of all nodes and, threfore, of all levels and > > of all finite trees does exist in the original set theoretic meaning. > > But since in each WW tree every path is eventually constant, in the > union of all such trees every path is still eventually constant, and no > path in that union can have both infinitely many 0 branchings and > infinitely many 1 branchings. > > Thus WM's "union" of WW trees is not the complete binary tree he claims, > and the countability of WW paths does not imply the countability of the > set of apths of the complete infintie binary tree. > > So WM fails again! > > If the infinite union of finite trees, T(oo), is the complet tree, T, concerning the set of nodes and edges, then both are identical with respect to all properties including all paths too. There is simply no further parameter to fix. If you do not accept this identity, then further discussion between us about this topic is not meaningful. Then you may maintain ZFC and may say that it is free of contradictions, but you must tolerate that a unique structure may yield different results, i.e. is not unique. i do not tolerate it. But it may be a matter of personal taste. Regards, WM
From: mueckenh on 25 Jan 2007 04:30 On 24 Jan., 21:05, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > Franziska Neugebauer schrieb: > >> You have to choose whether you want to see a tree as > > >> a) the standard tree, T = (M, E) > >> b) the "Virgil"-tree. T = (set of paths having certain properties). > > >> Which definition do you choose? > > > I defined the trees which I use by nodes and edges.[...] > > ,----[ WM in <1169545581.873125.47...(a)q2g2000cwa.googlegroups.com> ] > | It is defined. It is the tree which contains the root node and if it > | contains the tree with n levels, then it contains the tree with n+1 > | levels. > `---- What is wrong? > > >> If so then the set of paths of G is cardinally greater than the > >> std-union of sets of paths of the trees in the set of all finite > >> trees V*. > > > The set of paths of a tree is solely given by its nodes and edges.In Virgil's trees a tree is given by a set of paths. > > > Same nodes and edges result in same paths. Everything else is > > nonsense. > 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. If T(oo) = T concerning the set of nodes and edges, 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. > > [...] > > >> 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. > > > 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. 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. > > ,----[ <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. >Counting occurs in the finite.What you do _is_ "counting over to the infinite" By no means! Every n in N is finite. > 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. > > >> 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, i.e., that one of the tree contains paths which the other does not contain, then we should stop. I am not willing to waste my time with exlicit nonsense. (Set theory hitherto was only implicit nonsense.) > > >> > If you try to construct the tree with n levels, do you fail at some > >> > number of levels? No. Therefore the union is defined for every n. > > >> But not for V*. Recall: V* = { T(n) | n e omega } is the infinite set > >> of all finite binary trees. > > > It is a set which can be formed if the set of all n exists.In ZFC omega is the set of all natural numbers and it exists. If you > assume omega not to exist you can hardly show a contradiction in ZFC. If you assume omega to contain more than all natural numbers, then you are in error. Regards, WM
From: mueckenh on 25 Jan 2007 04:47
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.) The structure of my trees is given. Therefore even the edges are not necessary, but a guide for the eye. 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. > > > >> 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. > > > 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. 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. > 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, 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. Then you may maintain ZFC and may say that it is free of contradictions, but you must tolerate that a unique structure may yield different results, i.e. is not unique. My goal is reached. Regards, WM |