Cantor Confusion
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From: mueckenh on 17 Jan 2007 11:27 Franziska Neugebauer schrieb: > > You can read what _mathematically_ relevant in > > ,----[ http://en.wikipedia.org/wiki/Simple_path ] No. Just what is mathematically relevant, is not yet given there. Regards, WM
From: William Hughes on 17 Jan 2007 12:28 mueckenh(a)rz.fh-augsburg.de wrote: > 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. > I will switch to your defintion of a tree. A tree consists of a set of nodes, and contains all paths that the nodes define. > Of course not. The complete tree T1 contains all paths. T1 is not a tree. > > > > 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. No node is missing. The path is missing. T1 is not a tree. The fact that the nodes exist does not mean that the path exists. > > > > > > > > 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? T1 does not have the property that given a set of nodes in T1, the path through those nodes is in T1. > > > > 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. If you define at tree as a set of paths, then it is trivial to find two different trees with the same nodes. If you change the definition of tree to have a tree defined by its nodes, then two different trees cannot have the same nodes (DUH!). However, once you change the definition, T1 is not a 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. As defined by you A is not a tree. A is a set of paths. A contains the nodes M but does not contain every path defined by the nodes M. > > > > > > > > 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? We can tell what number they represent. However, we cannot tell whether this string of digits is a member of the tree (old definition). Using your definition of a tree there is no such thing as a tree that is not node-complete, but on the other hand, T1 is not a tree. > > > > 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. No. Any tree that contains T1 must contain an infinite number of nodes. Any tree that contains an infinite number of nodes must contain an infinite path. T1 does not contain an infinite path. T1 is not a tree. > It is an infinite complete tree. It cantains all paths > which exist outside of the world of Harry Potter. No, the tree that contains all paths is T2. T2 is a tree, but T2 is not T1. > > > > > > > > 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. So? My point is that T1 is not the tree that you showed. Why is showing another tree that is not T1 relevant? > > > > > 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? The path. Only if we have a tree can we conclude that the path exists from the fact that the nodes exist. We do not have a tree. > > > > 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? Piffle. A finite path certainly exists. But L_D is not just any finite path. L_D is a finite path that contains every element of the (potentially) infinite path 0.111... L_D does not exist. >But 0.111... does exist. Yes (in your terms the potentially infinite sequence 0.111... exists). >In the tree T1. No. T1 is not a tree. 0.111.. is a (potentially) infinite path. T1 does not contain a (potentially) infinite path. > > > > > > > > > > > > > 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? Yes, and by the definition give by Cantor it does exist. >Whethr there are 2^aleph_0 numbers or not is a matter of > definition? No. The defintion of Cantor's diagonal is used in one proof of the the fact that there are 2^aleph_0 real numbers. There are other proofs that do not use this 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 T1 is not a tree. You say to the set of paths T1 >" you must contain any path > defined by its nodes". Waht happens? You add a lot of paths to T1. > Then if I say to the tree T1 T1 is not a tree. You say to the set of paths T1 > "you > must not contain any path defined by its nodes". What happens then? T1 cannot contain any paths. - William Hughes
From: Virgil on 17 Jan 2007 14:40 In article <1169029643.505329.81220(a)38g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Han de Bruijn schrieb: > > > stephen(a)nomail.com wrote: > > > > > > All you have proven is that you do not know what the Continuum Hypothesis > > > is. > > > > LOL! Should I add a dozen smileys ? > > :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) > > Add several dozens! > > The problem with your example is that set theorists distinguish very > sharply between 2^omega and 2^aleph_0. On the other hand, they use > omega instaed of aleph_0. Cantor already did so. > There are some ways to show that there is no difference between 2^omega > and 2^aleph_0. Compare my recent arguing which forces set theorists to > assert that two binary trees with identical nodes contain different > paths. That is just saying that a set (of nodes) can have different subsets (paths). We start by assuming that all trees and all paths issue from a common root node. We may embed every such finite binary tree in a complete infinite binary tree by extending each finite path infinitely in the opposite direction of its last branching, so each such infinite path is "eventually constant", and the original path can be recovered from the infinite path. Then each finite binary tree extends to an /incomplete/ infinite binary tree which is a proper subtree of the /complete/ infinite binary tree. (A tree is /complete/ if and only if every non-leaf node has the same number of child nodes as all others). Then every finite path in every finite subtree corresponds to a unique infinite and eventually costant path in the complete tree, and every finite tree to an incomplete infinite tree which is a subtree of the same compete infinite binary tree. The 'union' of all such incomplete infinite binary trees will necessarily be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the entire tree. Since every path in every one of the incomplete infinite binary trees in the union is 'eventually constant', every path in the union of all those trees will also be eventually constant. But "most" paths in a complete infinite binary tree are NOT eventually constant, but have infinitely many branchings in each direction both left branchings and right breanchings. But WM's union of trees cannot contain any path which has infinitely many branchings in each direction. One can then easily show that the set of paths in WM's union will be countable, but the set of paths in the complete infinite binary tree will be uncountable. Thus WM's argument that the union of finite binary trees generates a complete infinite binary tree is falsified.
From: Virgil on 17 Jan 2007 14:47 In article <1169030073.463923.305630(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > This is obviously not the union in the sense of set theory. Set > > theoretically a tree is a directed graph which is defined as an ordered > > pair (V, E). The union of two trees is in general not a tree at all. > > What I defined a sthe union of finite trees is the union in the sense > of set theory. The set theory union of trees is not a tree at all, but with suitable modifications one can construct something that works somewhat in the way WM intends. Unfortunately for WM, it disproves his claims: We start by assuming that all trees and all paths issue from a common root node. We may embed every such finite binary tree in a complete infinite binary tree by extending each finite path infinitely in the opposite direction of its last branching, so each such infinite path is "eventually constant", and the original path can be recovered from the infinite path. Then each finite binary tree extends to an /incomplete/ infinite binary tree which is a proper subtree of the /complete/ infinite binary tree. (A tree is /complete/ if and only if every non-leaf node has the same number of child nodes as all others). Then every finite path in every finite subtree corresponds to a unique infinite and eventually costant path in the complete tree, and every finite tree to an incomplete infinite tree which is a subtree of the same compete infinite binary tree. The 'union' of all such incomplete infinite binary trees will necessarily be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the entire tree. Since every path in every one of the incomplete infinite binary trees in the union is 'eventually constant', every path in the union of all those trees will also be eventually constant. But "most" paths in a complete infinite binary tree are NOT eventually constant, but have infinitely many branchings in each direction both left branchings and right breanchings. But WM's union of trees cannot contain any path which has infinitely many branchings in each direction. One can then easily show that the set of paths in WM's union will be countable, but the set of paths in the complete infinite binary tree will be uncountable. Thus WM's argument that the union of finite binary trees generates a complete infinite binary tree is falsified. > > > > So obviously you mean a different tree representation in terms of sets > > and/or a different definiens in the definition of "union". > > > > > Now apply your logic. > > > > To which definitons? > > I gave them. Look them up or leave it. I will not repreat them hre. We have looked, and they are wrong! it takes something like my definitions above to get a "union" of trees to be a tree at all. > > > > >> > It is the limit of a set of finite paths like N is the limit of all > > >> > of its finite initial segments. > > >> > > >> This should read: The binary representation of 1/3 is _not_ a member > > >> of the set of all finite paths _as_ N is _not_ a member of the set of > > >> all natural numbers. > > > > > > Correct. > > > > Aha. > > > > > One could add: Cantor's diagonal number is not a member of > > > the set of all finite initial segments of Cantor's diagonal number. > > > (And only such initial segments occur in Cantor's diagonal proof). > > > > Since no infinite string is a member of the set of all finite sequences > > of strings this is trivially true. So what? > > 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.) So that WM's EIT is like the successor of omega, containing all the members of omega together with omega itself.
From: Virgil on 17 Jan 2007 14:52
In article <1169030565.542761.57580(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > The trees are sets of nodes. > > > > Trees usually consist of a set of nodes _and_ a set of *edges*. The set > > of paths in a particular tree is defined only if _both_ of the sets are > > given. > > > > Every edge is related to a node. The set of nodes is the set of ends of > edges united with the set of root nodes (which is only one). Edges can > but need not be used. Something must be used to determine which nodes are connected by an edge, and for each such connection, which is the parent and which is the child node in that connection. > One can use only nodes as well as one can use > only edges. Nodes with no further information does not give a tree. To do WM's "union" of trees being a tree we need something like: We start by assuming that all trees and all paths issue from a common root node. We may embed every such finite binary tree in a complete infinite binary tree by extending each finite path infinitely in the opposite direction of its last branching, so each such infinite path is "eventually constant", and the original path can be recovered from the infinite path. Then each finite binary tree extends to an /incomplete/ infinite binary tree which is a proper subtree of the /complete/ infinite binary tree. (A tree is /complete/ if and only if every non-leaf node has the same number of child nodes as all others). Then every finite path in every finite subtree corresponds to a unique infinite and eventually costant path in the complete tree, and every finite tree to an incomplete infinite tree which is a subtree of the same compete infinite binary tree. The 'union' of all such incomplete infinite binary trees will necessarily be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the entire tree. Since every path in every one of the incomplete infinite binary trees in the union is 'eventually constant', every path in the union of all those trees will also be eventually constant. But "most" paths in a complete infinite binary tree are NOT eventually constant, but have infinitely many branchings in each direction both left branchings and right breanchings. But WM's union of trees cannot contain any path which has infinitely many branchings in each direction. One can then easily show that the set of paths in WM's union will be countable, but the set of paths in the complete infinite binary tree will be uncountable. Thus WM's argument that the union of finite binary trees generates a complete infinite binary tree is falsified. |