Cantor Confusion
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From: Andy Smith on 19 Jan 2007 09:13 >I never thought you did, and agree that Andy Smith >would have a _very_ hard time. > Poo, no I wouldn't have a problem with suggesting something like that - but I would have a problem demonstrating that it had the required properties on other than an informal level .... (can you do that?) -- Andy Smith
From: Andy Smith on 19 Jan 2007 09:25 With due rigor, I mean. Obviously when x is rational E some m such that m! x = integer, cos(m! pi x) = +/- 1, otherwise |cos()|<1 and |cos()|^(2*n) -> 0. But I would expect some hard time especially over whether it is reasonable to talk about m! as m->oo and whether lim n->oo {x^n} A |x|<1 is really 0.00.. and indeed whether that is the same as integer 0. In message <1zJ+zeMgINsFFwba(a)phoenixsystems.demon.co.uk>, Andy Smith <Andy(a)phoenixsystems.co.uk> writes >>I never thought you did, and agree that Andy Smith >>would have a _very_ hard time. >> >Poo, no I wouldn't have a problem with suggesting something like that - >but I would have a problem demonstrating that it had the required >properties on other than an informal level .... > >(can you do that?) -- Andy Smith Phoenix Systems Mobile: +44 780 33 97 216 Tel: +44 208 549 8878 Fax: +44 208 287 9968 60 St Albans Road Kingston-upon-Thames Surrey KT2 5HH United Kingdom
From: Franziska Neugebauer on 19 Jan 2007 09:37 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> >> { 1 } U { 1, 2 } = { 1, 2 } is simply true, but does not "hold". >> > >> > No? >> >> There is no free variable in "{ 1 } U { 1, 2 } = { 1, 2 }" >> for which it could hold for. It is not a statement form. > > Not an empty set allowed for in this case? A statement form (Aussageform) F(x) "holds" for a value x iff F(x) is true. Since { 1 } U { 1, 2 } = { 1, 2 } is not a statement form but a statement (Aussage) it cannot "hold for something". There is no free variable involved. > Above we see a statement in every-day language. This statement is > true, i.e., it holds in every-day language (although it will seldom be > expressed in every-day language). The context of your statement is: ,----[ <45af2ebb$0$97230$892e7fe2(a)authen.yellow.readfreenews.net> ] | > The union of {1} and {1,2} is {1,2}. This holds for initial | > segments of N as well as for finite trees, cut or as Trauerweide. | | { 1 } U { 1, 2 } = { 1, 2 } is simply true, but does not "hold". `---- So what you in "every-day-language" say is as meaningless as 1 = 1. This holds for initial segments of N as well as for finite trees .... It is meaningless. >> >> >> > If there is an infinite diagonal, however, then there is no >> >> >> > reason for the path 0.111... of the union of finite trees to >> >> >> > be finite. >> >> >> >> >> >> The question is whether 0.[01] is in the "union of all finite >> >> >> paths". >> >> > >> >> > If it exists, >> >> >> >> It exists and it is 0.[01]. >> > >> > Then it exists in the union of all finite paths. (Every segment of >> > 0.010101... exists there, so the whole number does.) >> >> Define union of all finite paths first. > > The union of all paths turning always right gives the path: 0.111... > This union does exist without any futher definition. > > But the union of all alternating paths does not exist? Again: Define union of all finite paths first. >> >> Define "union of all finite trees". We are currently talking about >> >> finite paths. >> > >> > I did. >> >> "Define" here means to formally define. > > The union of all paths turning always right gives the path 0.111... Define union _of_ _all_ _finite_ _paths_ first. > This and some other paths do exist in the tree. In which tree? > Where is the paths due to 1/pi? If it exists, formally defined, then > it exists as defined by the initial segments of the series of digits > or bits (Cantor's Fundamentalreihen). Then it exists in the tree too, > because the tree is nothing but a representation of real numbers. If > 1/pi does not exist, as I suspect, then it does neither exist in the > tree. Then, even the infinite tree does not exist. I suspect your thoughts are not well-ordered. >> You shall once and for all commit yourself to whether your want to >> argument with >> >> 1. Union of trees. >> 2. Union of nodes. >> 3. Union of edged. >> 4. Union of paths. >> > No. Each one will be applied where approriate. Otherwise you cannot > see the contradiction. Many thanks for this valuable information. >> >> > because for every end node n of a path, there is a path crossing >> >> > n. >> >> >> >> Please elaborate. >> > >> > For every node on level n of a path, >> >> You switch between trees and paths. > > and levels and edges and sets and subsets. If you are not familiar > with these things, please ask for further hints. I am familiar with these things. Hence I can distinguish between them. > Paths are subsets of the set of nodes. This is in general not true. To meaningfully speak of paths you either must define them as sequence of nodes oder at least as set of edges. F. N. -- xyz
From: Franziska Neugebauer on 19 Jan 2007 09:48 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > A path is a subset of the set of nodes. >> >> Since every subset of the set of nodes of a tree is a set of nodes >> you eventually say: "A path is a set of nodes". This is not the whole >> truth. >> >> A path _has_ a set of nodes (besides the information about the >> connectivity of these nodes). >> >> > That does not mean that every subset of the set of nodes is a path. >> >> Actually no set of nodes _is_ a path. > > In a well defined tree, as my CT and my WWT are, every path is an > ordered set of nodes. An ordered set of nodes is not a set of nodes but a set of nodes _together_ with its order relation. Until now we have three ways to define "path": a) a sequence of nodes (vertices), b) a set of edges, c) an ordered set of nodes (vertices). In each case a path is not a plain set of nodes. Hence I am maintaining my critique of your claim A path is a subset of the set of nodes. > It is a subset of the complete ordered set of nodes which is the > infinite tree. I would like to read that in formal notation. F. N. -- xyz
From: Dik T. Winter on 19 Jan 2007 10:07
In article <1169211532.179686.171640(a)51g2000cwl.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > The paths in a tree are completely defined by the sequences of nodes > > > (or edges) which can be followed to an end in CT or without an end in > > > their union as well as in WWT. > > > > In what way does that contradict what I wrote? I was talking about sets > > of paths and their union. What you write is completely unrelated to that. > > You say the set of paths in the union tree is ot the same as the set of > paths in the complete tree. That is not what I say. I say that the set of paths in the union or complete tree is not the same as the union of the sets of paths in the finite trees. > According to set theory (including the axiom of choice) a countable > union of countable sets is a countable set. The set of paths in the > union tree T1is merely a countable union of finite sets, And that is wrong. The set of paths is *not* the union of the sets of paths in the finite trees. And that is easy to prove. The path 0.010101... is not in that union of sets of paths because it is in *none* of the sets of paths, so it also can not be in the union of sets of paths. > An index denotes the level to which a node belongs. The union of all > indexes of nodes of finite paths is the union of all initial segments > of natural numbers > {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n}... U ... = {1, 2, > 3, ...}. > This is also the set of all last elements of the finite segments, i.e., > it is the set of all natural numbers N. This is the set of all indexes > - there is no one left out. With respect to this observation we > examine, for instance, all finite paths of the tree T1 which always > turn right: 0.1, 0.11, 0.111, ... If considered as sets of nodes, their > union is the infinite path representing the real number 0.111... = 1. Right. > >From this we can conclude that also every other infinite path belongs > to the union T1 of all finite trees. Yes, it is in the union of finite trees. But *not* in the union of sets of paths in finite trees. > We see, the trees T1 and T2 are identical with respect to all nodes, > all edges, and all paths (which would already have been implied by the > identity of nodes and edges). But the set of all paths is countable in > the tree T1 and uncountable in the same tree T2. It is not countable, you have not given a proof. > > > > Eh? There is a whole collection of different sets of paths in a > > > > tree with all nodes (assuming the definition of tree as I stated > > > > above, that is the pair [set of nodes, set of edges]). Of course > > > > it is silly to suspect different complete sets of paths. > > > > > > So it is. > > > > And there is not. > > ??? The union of the sets of paths in the finite trees is *not* the complete set of paths. > > > > > Therefore the only conclusion is that > > > > > the path 0.010101... does not exist. > > > > > > > > Wrong. > > > > > > If it exists, then it exists in every tree with all nodes and edges. > > > > Yes, and I never contradicted that. > > Then it exists in the union tree - tgether with all paths of same > length. But I am not contradicting that. Again, I state that it is not in the union of sets of paths in finite trees. There is a huge difference. > > > > Again, I do not state that. What I state is that that path is > > > > (with the definitions above) in the complete tree. > > > > > > It is in every tree which contains all nodes and edges (as defined by > > > me), because there is no end when we follow the path always swithing > > > betwen 0 and 1. > > > > Yes, so what? I never said that was not the case. > > Which paths then are you missing in the union tree? I am not missing something in the union tree, I am missing something in the union of the sets of paths. > > > Is it or part of it outside of this union? > > > > This is a silly question. Something is an element of a set or somthing > > is not an element of a set. You can not have something being partly an > > element of a set. > > As there is no last node of a path, we can nly investigate subsets of a > path. Irrelevant. Something is in a set or something is not in a set. Something is not partly in a set. > > > You say it exists. It exists not in the union. Where does it exist? > > > > Outside the union. Let's illustrate. Call the diagonal of the triangle > > of order n: d_n. The union of the sets of diagonals of the triangles is: > > {d_1, d_2, d_3, d_4, ...} > > the diagonal d of the infinite tree is not in it. > > It is the union. Yes, so it is not *in* the union. > So if we have the union of all digits, then we have > the diagonal. The union of the digits is *not* the union of the sets of diagonals. > > Consider a tree of level 1 and a tree of level 2. The tree of level 1 > > contains only paths going through two nodes, the tree of level 2 only > > paths going through three nodes. So none of the paths in the level 1 > > tree is a path in the level 2 tree. So the union of the sets of paths > > contains both all paths of the level 1 tree and all paths of the level > > 2 tree. The set of paths of the level 2 tree contains only the paths > > of the level 2 tree. > > This is a wrong approach. (It is insufficient to observe the problem.) It is the only possible approach when you consider unions of sets of paths. Because that is the way the union of two sets is *defined*. A something is an element of the union of a collection of sets if it is an element of one of the constituent sets. Something is *not* en alement of the union of a collection of sets if it is not an element of *any* of the constituent sets. > > There are no nodes or edges in the union. The union is a union of sets > > of paths, so it only contains paths. How you manage to get nodes and > > edges as elements of a set of paths escapes me. > > The elements of our sets are nodes. The paths are subsets. If you do > not observe this fact, you cannot find a contradiction. You are using a countable union of *sets* of paths. If you do not observe this fact you are talking nonsense. That is: you state the set of paths is the countable union of finite sets. A set of paths has only one kind of subsets: sets of paths. So if it is a union it is the union of sets of paths. The set of paths is *not* the set of nodes. > > Wrong. That is not how set theory works. If every subset of a path is > > in the union, then the union contains (as elements) a whole lot of > > subsets of paths. But not the complete path. Consider the following set: > > {{1}, {2}, {3}, ...} > > this set does *not* contain (as element) {1, 2}. > > Your wrong approach again. By subsets linear subsets (initial segments) > are meant. By subset of a set of paths only sets of paths can be meant. Nothing else. That each path in turn is a set of nodes does not matter at all. And when you formulation the union of a collection of sets and get a set of paths, each set in the collection is a set of paths. > > > Every initial segment of a sequence in the union of trees. All initial > > > segments are countable. > > > > Indeed, all (proper) initial segments are countable. But this does show > > nothing. > > See above. See above. > > > Correct. All paths are in the union of all finite trees. > > > > And I did never contradict this. But not all paths are in the union of > > the sets of paths from the finite trees. There is quite some distinction. > > And in your proof of the countability of paths you assume the second. > > See above. Your examples with {1}, {2}, and {1,2} show a fundamental > misconception. You contradicting me here shows a fundamental misconception. You state that the set of paths is the countable union of finite sets. As each of those finite sets must be a subset of the set of paths, it must itself be a set of paths. Nodes and edges are irrelevant here. Consider, again, the trees of level 1 and 2. Consider a path as the union of initial segments. And consider terminating paths for simplicity, the same works when you have leave edges, but they complicate the matter a bit. So the paths in the level 1 tree are: {0., 0.0} and {0., 0.1}, the paths in the level 2 tree are: {0., 0.0, 0.00}, {0., 0.0, 0.01}, {0., 0.1, 0.10} and {0., 0.1, 0.11}. So the set of paths are: L1 {{0., 0.1}, {0., 0.1}} L2 {{0., 0.0, 0.00}, {0., 0.0, 0.01}, {0., 0.1, 0.10}, {0., 0.1, 0.11}} As none of the elements of L1 is an element of L2 and reverse, their union is a set of six paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |