Cantor Confusion
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From: Franziska Neugebauer on 17 Jan 2007 09:40 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> > What I defined as the union of finite trees is the union in the >> > sense of set theory. >> >> I remember that you refused to use usual graph theoretical approach >> where a tree is a directed graph. A graph is an ordered pair of a set >> of nodes _and_ a set of edges. Please show that in general a union of >> trees (ordered pairs of sets of nodes and sets of edges) is a tree. > > I do not assert that this is so in general. Therefore I will not prove > it. Then please don't call it "union". > What I said is that the union of the trees with n levels and the > tree with m >= n levels is the tree with m levels. So your "union" is merely a kind of "maximum". For there is no max. element n e N the "union of all finite trees" is undefined in the first place. > This is valid for > the levels, for the nodes, for the edges, and for the paths of trees > like the following with n = 2 levels. > >>> > 0. >>> > / \ >>> > 0 1 >>> > / \ / \ >>> > 0 1 0 1 >>> > | | | | >>> > 0 1 0 1 >>> > | | | | >>> > 0 1 0 1 >>> > ........................... > > >> > >> > The Equilateral Infinite Triangle (EIT) contains all lines which >> > can be enumerated by natural numbers: >> > >> > 1 0.1 >> > 2 0.11 >> > 3 0.111 >> > ........... >> >> This is simply a(n infinite) list (sequence) of unary >> representations. > > The diagonal has an infinite number of digits. So what? ,----[ context restored ] | > In addition it contains the complete diagonal 0.111... [(1)] | | Since Sigma^+ does not contain an element "1..." your "EIT" sequence | does not contain it as element either. So your claim is _wrong. | | > (Otherwise, Cantor's diagonal proof would fail.) [(2)] | | 1. Non sequitur. | 2. Cantor's proof does not "fail" in that case cause 0.1... is not | in the "EIT"-list (i.e. there is no pair (n, 0.111...) e EIT, | n e N). `---- >> > (Otherwise, Cantor's diagonal proof would fail.) > >> 2. Cantor's proof does not "fail" in that case cause 0.1... is not >> _in_ the "EIT"-list (i.e. there is no pair (n, 0.111...) e EIT, >> n e N). > > I said *otherwise* Cantor's proof would fail. "Otherwise" means: If > there is no infinite diagonal. Your claimed (please take a look into the restored context): 1. Sigma^+ contains "1..." (1) 2. Do you by "A otherwise B" mean 2a. ~A -> B 2b. A or B. 2c. ? With "A" being (1) and "B" "Cantor's diagonal proof would fail". What we know is, that i. (1) is true. There is an infinite diagonal (by definition). ii. Cantor's proof does not fail in that case, since 0.111.. is not a member in the "EIT"-sequence. Hence B is true. In case that (2) has meaning 2a. your statement is false. Proof: ~A is true B is false (true -> false) is false. q.e.d. In case that (2) has meaning 2b. your statement is false, too. Proof: A is false. B is false. (false or false) is false. q.e.d. > 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". F. N. -- xyz
From: Franziska Neugebauer on 17 Jan 2007 09:46 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> > Every edge is related to a node. >> Every edge is actually related to _two_ nodes (vertices). > > And sometimes they are drawn with black ink while the nodes are red. > Sometimes colours are reversed or even different colours are used. > Relevance? I had the suspect that you confuse your sketches on a sheet of paper with what mathematicians call a tree. >> > The set of nodes is the set of ends of edges united with the set of >> > root nodes (which is only one). >> >> The "set of ends of edges" (united with a set of roots or not) does >> not _constitute_ the tree. Nonetheless this union may perhaps _equal_ >> the set of vertices. >> >> > Edges can but need not be used. >> >> Whether they "need" to be "used" is irrelevant. Relevant is: > > that definition which I gave for that kind of trees which I use in my > proof. You can read what _mathematically_ relevant in ,----[ http://en.wikipedia.org/wiki/Simple_path ] | In graph theory, a path in a graph is a sequence of vertices such that | from each of its vertices there is an edge to the next vertex in the | sequence. `---- As I have mentioned in my last post: You have not given a definition of a union of all finite trees. F. N. -- xyz
From: mueckenh on 17 Jan 2007 10:56 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > > > > If there is no difference in terms of nodes, why does T1 not contain > > > > all possible paths? > > > > > > Because the paths define the nodes, not the other way round. > > > > Wrong. My trees are defined by nodes. > > In this case T1 is not a tree. There is no set of nodes > which contains all finite paths but does not contain > an infinite path. Of course not. The complete tree T1 contains all paths. > > T1, the union of all finite trees, contains only finite paths. You > cannot make > T1 contain an infinite path by playing with > definitions. Then show me which node of the path 0.010101... is missing in T1. > > > > > > Given any tree (set of paths) we can add paths to > > > it to make it node-complete. > > > > > How can we add? How can we add a path??? > > > > > T1 is not node-complete. T2 is node-complete. > > > To make T1 into T2 we have to add paths to > > > T1 to make it node complete. > > > > That is a nice story. Can you say what has to be done in order to make > > a tree node complete? > > Using your definition of tree, nothing. Using your > definition, there is no such > thing as a tree that is not node-complete. > However, using your definition, T1 is not a tree.# No? What is missing? You know that tere are finite trees in mathematics? > > If you define a tree to be a set of paths, then > two different trees can have the same nodes. Not if all paths go downwards as regularly as in my tree. > > Start with one set of paths A, these > paths contain nodes M. The nodes M contain > the set of paths A, and as well they contain the > set of paths B. Add the set of paths A to > the set of paths B to get the set of paths C. > C contains the same nodes as A, but C is > not the same set as A. Simply consider the trees as defined by me. > > > > > What must be done to make Cantor's matrix non-node-complete (i.e., to > > remove the diagonal without changing the hard ware)? Or what has to be > > done to make any tree non-node-complete, for instance a complete tree > > or a tree which contains 0.010101... but not 0.010110110...? How can we > > recognize whether a tree is (partially) node-complete? > > We cannot. We cannot recognize from a string of digits, what number they represent? > > If you define a tree to be a set of paths then > just looking at the nodes will not tell you which paths are > in the tree. > > If you define a tree to be a set of nodes, and > any path through the nodes is in the tree, then looking > at the nodes will tell you what paths are in the tree > but T1 is not a tree. It is a tree. It is an infinite complete tree. It cantains all paths which exist outside of the world of Harry Potter. > > > > > Try to find the difference by following simple construction: > > > > The Equilateral Infinite Triangle (EIT) contains all lines which can be > > enumerated by natural numbers: > > > > 1 0.1 > > 2 0.11 > > 3 0.111 > > ........... > > > > In addition it contains the complete diagonal 0.111... (Otherwise, > > Cantor's diagonal proof would fail.) > > > > The union of all finite binary trees contains all levels which can be > > enumerated by natural numbers: > > > > Correct > > > 0 0. > > / \ > > 1 0 1 > > / \ / \ > > 2 0 1 0 1 > > ............................... > > > > However, this is not T1. Using your definition of tree, > the union of all finite trees is not a tree. Look at the original definition: 0. / \ 0 1 / \ / \ 0 1 0 1 | | | | 0 1 0 1 | | | | 0 1 0 1 ............................ This is a 1-level tree. > > > But it does not contain the complete path 0.111... (Otherwise Cantor's > > proof of the uncountability of the real numbers was contradicted). > > Correct. A union of finite paths does not contain an infinite > path. Note, however, that the union of finite paths, does contain > every node of the infinite path. What is then missing to make the infinite path exist? > > Let L_D be the single finite path that contains every element > of the (potentially) infinite path 0.111... > > L_D is a finite path Then you should be able to say where it ends. That is the definition of "finite". > 0.111... is a potentially infinite path > > L_D does not exist. No? A finite path does not exist? But 0.111... does exist. In the tree T1. > > > > > > > > Now, the path 0.111... in fact would look like a diagonal > > > > 0 0. > > 1 1 > > 2 1 > > 3 1 > > ................. > > > > As noted the EIT contains every node needed to make > up the diagonal. Whether the EIT contains the diagonal > is a matter of definition. Interesting. Whether Cantor's diagonal exists is a matter of definition? Whethr there are 2^aleph_0 numbers or not is a matter of definition? > > > > > Are the different notations "line" and "level" the reason for the > > difference? > > > > No. The difference is whether you say that a tree must > contain any path defined by its nodes. The difference? I say to the tree T1 " you must contain any path defined by its nodes". Waht happens? Then if I say to the tree T1 "you must not contain any path defined by its nodes". What happens then? Where is a difference? Regards, WM
From: mueckenh on 17 Jan 2007 11:16 Dik T. Winter schrieb: > > > > > > > > 0. > > > > / \ > > > > 0 1 > > > > / \ / \ > > > > 0 1 0 1 > > > > | | | | > > > > 0 1 0 1 > > > > | | | | > > > > 0 1 0 1 > > > > ........................... > > This finite tree I call: Trauerweide > Ok. But this does not help either. The complete tree contains the > path 0.0101010101...; none of the 'finite' trees contains that path. > So that path is *not* in the union of the sets of paths of the finite > trees, while it is in the union of trees. So you say 0.0101010101... is in the union of finite trees while ohers say it is not? It cannot be in the union of the finite trees, they say, because it is not in one of the finite trees. But *all* paths are in the union of the finite trees, because it is simply silly to suspect different sets of paths in a tree with all nodes. Therefore the only conclusion is that the path 0.010101... does not exist. > > > > > The answer is: We can use sets of levels, sets of edges, sets of nodes > > or sets of paths. Look at the "Trauerweide". > > Yes. But still the union of the sets of paths in the finite trees is not > the set of paths in the complete tree. There are paths in the complete > tree that are in *none* of the finite trees and so not in the union of those > sets of paths. So you say. If you were right, there should be some indication for the difference. What makes the path 0.010101... be in a tree with all nodes but not in a tree with all nodes? > > > > Which digit is missing? > > > > > > None. But all sets of diagonals of finite squares contain only finite > > > diagonals, so there can not be an infinite diagonal in their union > > > (which is a set of infinite size). > > > > So there cannot be an infinite diagonal at all? > > I do not state that. I state that in the union of the sets of diagonals > of the finite squares only finite diagonals do exist (infinitely many > of them). This does *not* prevent an infinite diagonal to exist. What can we do to switch its existence on and off? > > > The Equilateral Infinite Triangle (EIT) contains all lines which can be > > enumerated by natural numbers: > > > > 1 0.1 > > 2 0.11 > > 3 0.111 > > ........... > > > > In addition it contains the complete diagonal 0.111... (Otherwise, > > Cantor's diagonal proof would fail.) > > Right. But the complete diagonal is not in the union of the sets of > diagonals of the finite triangles. Where is it? Or is it nowhere? > > > The union of all finite binary trees contains all levels which can be > > enumerated by natural numbers: > > > > 0 0. > > / \ > > 1 0 1 > > / \ / \ > > 2 0 1 0 1 > > ............................... > > > > But it does not contain the complete path 0.111... (Otherwise Cantor's > > proof of the uncountability of the real numbers was contradicted). > > It does contain that path (according to your definition above). Yes, that is obvious. But, for a moment, switch to the cut trees, please. Does the union of all cut trees contain the paths 0.111...? You would say no. > But > the union of the sets of paths in the finite trees is not the set of > paths in the infinite trees. The first does not contain the path > 0.01010101..., but it is in the second. If this path exists, then it must be constructible from all the nodes present. There is a bit, 0 or 1, for every index n in N. That is not sufficient? > > > > > There are more limits than sequences? > > > > > > Eh? Why do you think so? Each sequence has only a single limit (when the > > > limit is properly defined and when it exists). > > > > Correct. > > So why your question? There are only countable many sequences. > > > > > What we need is: All paths > > > > in the union tree are countable. All possible nodes are in the union > > > > tree. > > > > > > Yes, you desperately need it. I do not need it, but you do. > > > > The countablity of all paths in the union of finite trees is a theorem > > of set theory. > > No, it is not. Prove it. You use the *false* assumption that the > union of the sets of paths in the finite trees is the set of paths > in the complete tree. It is not because the first does not contain > the path 0.01010101... . I use the fact that the union of all nodes contains the nion of all paths. > > > > > A path can only then differ from all paths of the union tree if > > > > there is another node (or if it is running in other directions --- try > > > > that argument?). > > > > > > What is the argument here? I think you are back to the old argument, that > > > is not valid either. > > > (1) For each pair of paths there is an edge were they differ. > > > > in the union of all finite trees. Accepted. > > > > > (2) For each path there is no edge where it differs from all other paths. > > > > in the union of all finite trees. Accepted. > > But you state above "A path can only then differ from all paths of the > union tree if there is another node". What do you mean with that? I mean that in an infinite tree (= union of all finite trees) with all nodes, there is a set of paths. Paths which are not therein, must indicate why they are not therein - not be a spell but by a bit (or node). > (1) and (2) together state that while all paths are different, there > is *no* single edge that makes a single path stand out from all other > paths. But paths which are not determined by the set of all nodes must stand out by some indication, or they are determined by the set of all nodes - or they are existing in imagination only. Regards, WM
From: mueckenh on 17 Jan 2007 11:25
Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> > What I defined as the union of finite trees is the union in the > >> > sense of set theory. > >> > >> I remember that you refused to use usual graph theoretical approach > >> where a tree is a directed graph. A graph is an ordered pair of a set > >> of nodes _and_ a set of edges. Please show that in general a union of > >> trees (ordered pairs of sets of nodes and sets of edges) is a tree. > > > > I do not assert that this is so in general. Therefore I will not prove > > it. > > Then please don't call it "union". > > > What I said is that the union of the trees with n levels and the > > tree with m >= n levels is the tree with m levels. > > So your "union" is merely a kind of "maximum". For there is no max. > element n e N the "union of all finite trees" is undefined in the first > place. 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. > > > This is valid for > > the levels, for the nodes, for the edges, and for the paths of trees > > like the following with n = 2 levels. > > > >>> > 0. > >>> > / \ > >>> > 0 1 > >>> > / \ / \ > >>> > 0 1 0 1 > >>> > | | | | > >>> > 0 1 0 1 > >>> > | | | | > >>> > 0 1 0 1 > >>> > ........................... > > > > > >> > > >> > The Equilateral Infinite Triangle (EIT) contains all lines which > >> > can be enumerated by natural numbers: > >> > > >> > 1 0.1 > >> > 2 0.11 > >> > 3 0.111 > >> > ........... > >> > >> This is simply a(n infinite) list (sequence) of unary > >> representations. > > > > The diagonal has an infinite number of digits. > >> > (Otherwise, Cantor's diagonal proof would fail.) > > > >> 2. Cantor's proof does not "fail" in that case cause 0.1... is not > >> _in_ the "EIT"-list (i.e. there is no pair (n, 0.111...) e EIT, > >> n e N). > > > > I said *otherwise* Cantor's proof would fail. "Otherwise" means: If > > there is no infinite diagonal. > > Your claimed (please take a look into the restored context): Not necessary. > > i. (1) is true. There is an infinite diagonal (by definition). But there is no infinite path by definition? > > > 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, then it must be in the union, because there is nothing remaining outside of the union of all finite trees and all finite paths which could extend any of its paths. The union of all finite paths is an infinite path, because for every end node n of a path, there is a path crossing n. Regards, WM |