Cantor Confusion
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From: Virgil on 21 Jan 2007 15:48 In article <1169375378.815811.212630(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > UT is not a tree. > > > > > > > > > > Call it as you like. Simply call it T1. (If I speak of "tree" blow, > > > simply read "eert".) It may be what you like. In any case it is the > > > same by nodes, edges > > > > Yes > > > > > and paths as T2. > > > > No. > > The directions of the paths are the same in T1 and T2. If you insist on > a difference, then it can only result from the length of the paths. Two paths will be different it there is some node from which one branches in one direction and the other does not branch in the same direction (it may branch in another direction or it may have that as a terminal node). > > in T1 are not the same as the paths in T2. > > > > T1 contains only finite paths. > > This is in fact the crucial question. If you are right, then I have > been working in vain for four years. EM has been working in vain for years regardless of the outcome. > If you are wrong, then thousands > of mathematician have been working in vain for 130 years. WM is not competent to pass judgement on mathematicians. > Two trees which are identical by all nodes and edges are identical by > all paths. For finite sets, a tree is uniquely determined by its set of nodes and set of edges. This is false for infinite trees as the following demonstrates: Let Tc be the complete infinite binary tree, whose set of paths correspond to the complete set of all functions from N to {0,1}. Let Ti be the infinite tree whose set of paths are limited to those which are "eventually constant", corresponding to those functions from N to {0,1} for which either {n e N: f(n) = 0} or {n e N: f(n) - 1} is finite. One can easily see that the sets of nodes are the same for both trees and the sets of edges is also the same for both trees but the sets of paths are quite different, the set of paths of Ti being a proper subset of the set of pants of Tc. Furthermore, the set of paths of Ti is countable but the set of paths of Tc is not. > T1 as the countable union of all finite sets of finite paths contains > only a countable set of finite paths. > T1 = T2. Wrong! See above!
From: Virgil on 21 Jan 2007 15:58 In article <1169376257.114375.24450(a)38g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Trees which are identical with respect to nodes and edges cannot have > > > different *paths*. > > > > While that may be true for finite trees, it is demonstrably false for > > infinite trees. > > > > This limited infinite tree contains every node and every edge that the > > complete binary tree contains, > > If there are all nodes and all edges but only some of the paths: what > prevents the other paths from being there? The rule which excludes them. Does WM claim that the object that I have described, whose paths are all "eventually constant" is an infinite binary tree? If so let his demonstrate what tree property it lacks. > > > but it is clearly not the same tree. > > That is your wrong assertion. To assert it is wrong is WM's wrong assertion. > > > So WM is trivially and totally wrong in his claim above. > > > > It is WM's thoughtless assumption that every property of finite trees > > automatically also holds for infinite trees that leads him to make a > > fool of himself in this way. > > Personal insults will not really support your claims. But my explicit example of two infinite binary trees with the same nodes and edges but different sets of paths does support my claim and disproves WM's counterclaims. > > ======================= > > > The contradiction is in WM's false assumption that nodes and edges are > > eniugh. But unless (4) is applied and can be shown to generate ALL > > infinite paths, one cannot guarantee a complete infinite binary tree by > > any such unioning. > > Correct is: Two trees which are identical by all nodes and edges are > identical by all paths. That is false for infinite trees as per my example. By that example I have shown that the path of alternating left and right branchings belongs to one tree and not to another although both have the same nodes and the same edges. > T1 as the countable union of all finite sets of finite paths contains > only a countable set of finite paths. > T1 = T2. Repeating a falshood does not make it any less false. > > Regards, WM
From: mueckenh on 21 Jan 2007 15:59 William Hughes schrieb: > > The directions of the paths are the same in T1 and T2. If you insist on > > a difference, then it can only result from the length of the paths. > > Yes. T1 contains only finite paths. Which path is finite? Where does a path end? > T2 contains only > (potentially) infinite paths. > > > > > The paths in T1 are not determined by the nodes in T1. > > > > By its nodes and edges every path is determined. > > Yes, but the fact that these nodes and edges are in T1 > does not mean the path determined by these nodes > and edges in in T1. Even if the paths are in T1 this does not mean that the paths are in T1, I assume? > > > What are the nodes and edges for, in your opinion? > > Given a set of nodes, you can create paths. However, the > fact that a set of nodes M,is contained within a set of paths, S, > does not mean that every path that can be made by the nodes > M is in S. We do have "to create or make paths", in particular because you cannot define what we would have to do in order to make paths without changing nodes and edges. If there *are nodes* (connected by edges or my general pescription), then there *are the paths* corresponding to them. > > > Why do you think the EIT has an infinite diagonal? > > Because the diagonal of the EIT is definied as the (potentially) > inifinite sequence consistting of the ends of the lines in the EIT. > The diagonal of the EIT is not defined as the union of all > the finite diagonals of finite initial seqments of the EIT. Well, the union of ends cannot be larger than the union of all elements. Neertheless, that is irrelevant. You agree that the union of all line ends is an infinite path (the diagonal). In the union of al finite trees we have the uniuon of all ends of finite paths (= line ends in the EIT). > > > I say: The union of indexes of paths {1} U {1,2} U {1,2,3} U ... is an > > infinite index union {1,2,3,...} > > Correct and irrelevant. The union of indexes of paths is not > the same thing as the union of paths. Oh, yes. The paths are unions of nodes K(n) with K = 0 or 1 and n the index of the node. But there can be established a bijection between nodes and indexes by K(n) <--> n, can't it? > > corresponding to n infinite path. > > Incorrect. A path is not a set of indexes. > A path contains a set of indexes. > > > If > > you disagree, please prove your assertion. > > Where is the end of one path of T1? > > > > In the middle of the tree T2. Where is the middle of an infinite tree? T1 is not a tree. > The paths in T1 do not have to "go to the bottom". No? What about influence of gravity? > > Yes, given any path, P, in T1 > you can find a node in T1 such that you can extend path > P to path P1. Please give a path in T1 and find such a node. > This is irrelevant. There is nothing > in the definition of T1 that says a path in T1 cannot > be extended by a node in T1. Every path in T1 is, by definition, the union of an infinite set of finite sets of nodes {K(1)} U [K(1), (K(2)} U ... > > In mathematics a sequence is proven finite if and only if there is a > > natural number n_0 which not belonging to its domain. Otherwise a > > sequence is infinite. > > > > All irrelevant. A sequence is not its domain. > If you have a set of sequences S, you can take the union > of the domains, D, then construct T to be the set of sequences > whose domain is contained in D. However, you cannot conclude > that any sequence contained in T is contained in S. You can, if, by definition, all terms (which are defined) are 1. > > > > > > > > > > > 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, ...}. > > > > The union of domains {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} > > U ... is a domain {1, 2, 3, ...}. > > > > Correct, but a domain is not a sequence. Here, we are interested in domains only (length of paths). > > > > 0.111... is an > > > (potentially) infinite path. 0.111... can be seen as a limit > > > of a sequence of finite paths. It is not the union of a sequence > > > of finite paths. Union and limit are not the same thing. > > > (The union of {1/2},{3/4},{4/5} .... is not {1}). > > > > But the union of the intervals (0, n/(n+1)) has length 1. > > > > No. A union of intervals is not an interval. The interval is a set of points. The union of two sets contains all the elemets which are in at least one of the sets. This can be one interval. > The supremum of the > intervals > (0, n/(n+1)) has length 1. However, union and supremum are different > things. > > > And the union of singletons {1} U {2} U {3} U... U {n} U ... is not an > > infinite number omega? > > Yes it is. But as we are not discussing a union of > singletons, this is irrelevant. You are in error, but there are so much amusing assertions, please continue. Or are you trying a psychological experiment? > > > Two trees which are identical by all nodes and edges are identical by > > all paths. > > Correct, however, T1 is not a tree. But it is T1. And it is the same as T2 in every sensible respect. That is enough. > > > T1 as the countable union of all finite sets of finite paths contains > > only a countable set of finite paths. > > Correct. > > > T1 = T2. > ] > No. T1 and T2 contain the same nodes, but T1 does > not contain every path defined by its nodes. > T1 and T2 do not contain the same paths. Remember gravity. Regards, WM
From: Virgil on 21 Jan 2007 16:05 In article <1169376936.231595.281370(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > > >> Whether a _meaning_ of > > >> > > >> T(1) U T(2) U T(3) U ... (inf-u) > > >> > > >> exists does _not_ depend on such question. It depends on a definition > > >> you have to provide. So please answer the question, what (inf-u) > > >> means and prove that it exists. > > >> > > >> > This is connected > > >> > > >> Whether that is "connected" is irrelevant to a possible definition of > > >> (inf-u). > > > > > > That is not a true statement. Nevertheless: > > > > > > Definition of the infinite union of trees by induction: If the finite > > > tree with n levels exists, the finite tree with n+1 levels exist. The > > > tree with 1 level exists. > > > > By induction you only define (if at all) every _finite_ union > > > > T(1) U ... U T(n) > > That is completely sufficient as long as there is no upper bound, but n > --> oo (potential infinity). > > But if you, personally, think it is necessary, then simply use the > axiom of infinity. There is a set of all natural numbers, a set of all > levels, a set of all finite trees. But nothing that makes a union of finite trees into an infinite tree. While a union of sets is necessarily a set, trees are not merely sets, they have additional structure required, and the unioning of sets does not automate any appropriate amalgamation of the additional structures. WM is relying on his false assumption that an infinite tree is totally determined by its set of nodes and set of edges, even though he has been presented with two different infinite trees with the same sets of nodes and edges but differing sets of paths.
From: mueckenh on 21 Jan 2007 16:10
Franziska Neugebauer schrieb: > > nevertheless the union of {1,...,n} and > > {1,..., m} and the infinite union of all segments are defined. > > We are writing about trees. The trees have levels. For them we have the same as for the initial segments given above. > > >> | > >> | The questions is: How do you define U V_omega? V_omega is not in the union. The union is only over omega elements, but that is not a problem because it is covered by the axiom of infinity. Everything else is fine and finite. > > > > There is no question open. > > There is still an open question: How do *you* _define_ > > The infinite union of all finite trees? The finite trees have levels at finite positions. For the union of trees we have the same with respect to levels as for the initial segments given above. There is not the slightest difference. Regards, WM |