Cantor Confusion
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From: Virgil on 17 Jan 2007 15:05 In article <1169031082.277035.45310(a)v45g2000cwv.googlegroups.com>, 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. If there are no more than nodes, you don't have a tree at all, you have no more than a bare set of nodes. You must have in addition, for each pair of nodes, whether there is an edge connecting them , and if so, which node is the parent and which is the child. For something that works like a real "union of trees", 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 by deleting everything from the last change of direction onwards. Then each finite binary tree extends uniquely to an /incomplete/ infinite binary tree which is a proper subtree of the /complete/ infinite binary tree and the extension restricts uniquely to the original tree. (A tree is /complete/ if and only if every non-leaf node has the same number of child nodes as all others and all paths have the same finite length or are all infinite). Then every finite path in every finite subtree corresponds to a unique infinite and eventually constant path in the complete tree, and every finite tree to an incomplete infinite tree which is a proper subtree of a single compete infinite binary tree. The obvious �union� of all such incomplete infinite binary trees will necessarily be a subtree of the complete infinite binary tree. So far, I hope WM and I agree. The issue between WM and the rest of the world 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 (is not eventually constant). 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. As a matter of fact, Cantor's original diagonal proof was of the uncountability of the set of all binary sequences, which is exactly the same thing as the set of all paths in an infinite binary tree. 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 15:27 In article <1169032407.857382.32680(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > Since every path in every one of the restricted infinite binary trees in > > > > the union is eventually constant, every path in the union of all those > > > > trees will also be eventually constant. > > > > > > That would be correct for a finite union of finite trees. What is > > > correct for the finite case need not be correct for the infinite case. > > > > If every path in every tree in a union of trees has a certain property > > then every path in the resulting tree has that property. > > If every finite initial segment of N has the propery of being finite, > then the union of all initial finite segments cannot be an infinite > segment. N does not exist. False analogy. The correctly analogous statement is": if every member of every finite initial ssegemtn of the naturals is finite then every member of the union of thsoe finite initial segments is finite, i.e., every member of N is finite. > > > We have infinitely many finite sets of eventually constant paths. We > > form the set-union of all those sets of paths. And WM requires that the > > union of those sets contain something not in any of the separate sets. > > > > What does WM think a "union" of sets is? > > I think the union of all initial finite segments {1,2,3,...,n} is the > union of all natural numbers, namely N. Which, as above, has no infinite members. > > > > The proper definition of a union of sets is a set that contains x as a > > member if and only if at least one of the sets being unioned contains x > > as a member. Whether the union is of finitely many sets or infinitely > > many makes no difference to that definition. > > > > So whatever WM's idea of "union" is, it is not the mathematical union of > > sets. > > No? Then I will quickly withdraw my statements above and say: There > cannot be an infinite union N unless there is at least one infinite > number in N. But, alas, that would not be a finite number and, hence, > would not belong to N. Therefore there is no N. I had been advocating > this position already, once upon a time. So that WM reverts to his original error under pressure. We are only claiming that within ZF or NBG or similar axiom systems this holds. When WM can prove such systems are internally inconsistent, only then will we grant that a system using his axioms has any greater virtue.
From: Virgil on 17 Jan 2007 15:31 In article <1169032782.515373.159890(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <MPG.2016e6f7746f9f07989b49(a)news.rcn.com>, > > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > No. What I am using is this: > > > > 1) The union of all finite trees contains the union of all finite paths > > > > and this set is countable. > > > > 2) The union of all finite trees contains all nodes. > > > > 3) You cannot add another tree to the set of all finite trees without > > > > adding at least one node. > > > > (The paths are subsets of the set of nodes. Two different sets must be > > > > distinguished by at least one element.) > > > > 4) But there is no node to add. > > > > 5) Therefore the union of trees contains all possible paths. > > > > 6) We know that the set of all paths contained in this union is > > > > countable. > > > > > > How does 6 follow from 1-5? > > > > It doesn't, > > It follows from the theorem that a countable union of finite sets is > countable. > Every finite tree contains only a finite set of paths. But the union of those finite sets of finite paths is a set of finite paths, and the set of infinite paths having any one of those finite paths as initial section is uncountable. > The union of finite trees is countable. But irrelevant. > > Regards, WM
From: Virgil on 17 Jan 2007 15:33 In article <1169033684.287474.208720(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Dik T. Winter schrieb: > > > In article <1168962527.282591.33370(a)l53g2000cwa.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <1168955262.910310.48680(a)v45g2000cwv.googlegroups.com> > > > > mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > Please look at the paths in original definition: I'll try to draw it > > > again somewhat more suggestive: The finite tree from level 0 to level 1 > > > is: > > > > > > 0. > > > / \ > > > 0 1 > > > / \ / \ > > > 0 1 0 1 > > > | | | | > > > 0 1 0 1 > > > | | | | > > > 0 1 0 1 > > > ........................... > This finite tree I call: Trauerweide > > > > So it contains infinitely many nodes? You are again shifting your > > paradigm. > > And I would not think about that as a finite tree, only as a tree that > > splits at finitely many places. > > This was my first definition. The property "finite" belongs to the tree > because there are splittings only between level 0 and level n. This > shows that the union of paths can really be fomed. So your recent > objections are removed (therefore I do not answer your recent > contribtions.)For something that works like a real "union of trees", 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 by deleting everything from the last change of direction onwards. Then each finite binary tree extends uniquely to an /incomplete/ infinite binary tree which is a proper subtree of the /complete/ infinite binary tree and the extension restricts uniquely to the original tree. (A tree is /complete/ if and only if every non-leaf node has the same number of child nodes as all others and all paths have the same finite length or are all infinite). Then every finite path in every finite subtree corresponds to a unique infinite and eventually constant path in the complete tree, and every finite tree to an incomplete infinite tree which is a proper subtree of a single compete infinite binary tree. The obvious 'union' of all such incomplete infinite binary trees will necessarily be a subtree of the complete infinite binary tree. So far, I hope WM and I agree. The issue between WM and the rest of the world 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 (is not eventually constant). 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. As a matter of fact, Cantor's original diagonal proof was of the uncountability of the set of all binary sequences, which is exactly the same thing as the set of all paths in an infinite binary tree. 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 15:39
In article <1169038117.897324.208940(a)l53g2000cwa.googlegroups.com>, 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. 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. 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. WM's idealization many work for finite unions of finite and nested trees, but does not work for infinite unions, as it begs the question of how many infinite paths in an infinite tree have a given finite initial segment. WM assumes countably many and begs the question to conclude countably many. The following does not make any such assumption. For something that works like a real "union of trees", at least for infinite unions, 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 by deleting everything from the last change of direction onwards. Then each finite binary tree extends uniquely to an /incomplete/ infinite binary tree which is a proper subtree of the /complete/ infinite binary tree and the extension restricts uniquely to the original tree. (A tree is /complete/ if and only if every non-leaf node has the same number of child nodes as all others and all paths have the same finite length or are all infinite). Then every finite path in every finite subtree corresponds to a unique infinite and eventually constant path in the complete tree, and every finite tree to an incomplete infinite tree which is a proper subtree of a single compete infinite binary tree. The obvious 'union' of all such incomplete infinite binary trees will necessarily be a subtree of the complete infinite binary tree. So far, I hope WM and I agree. The issue between WM and the rest of the world 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 (is not eventually constant). 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. As a matter of fact, Cantor's original diagonal proof was of the uncountability of the set of all binary sequences, which is exactly the same thing as the set of all paths in an infinite binary tree. Thus WM's argument that the union of finite binary trees generates a /complete/ infinite binary tree is falsified. |