Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 8 Jan 2007 16:32 Dik T. Winter schrieb: > But in that case you have to properly define what you are doing. So pray > tell, how do you *define* the union of trees. See my last posting. > > > > You are dense. The well-ordering is given under the conditions of the > > > axiom V=L. > > > > THEN GIVE IT PLEASE, HERE AND NOW ! Simply reproduce it. > > I have not looked into it, but the axiom V=L immediately gives a > well-ordering, not only of R. Empty assertion! > > > > If the axiom is false the well-ordering is not given. > > > > How could it disappear after having existed? Who would force us to > > forget it, if it had been given? > > Did you look at that axiom? No, I think not. Of course I did not. The axiom may state the existence of a well-ordering but does not construct or define it (because it cannot be constructed or defined) > > > > What counts is only: There cannot be more unions than elements. > > Eh? As I understand it, there is only one union. But, what is the > relevance? > There is a union of paths for every path. There is a union of trees yielding the infinite tree. Every finite tree reaches from the root to a level n. The union contains every level which is a natural number. Every node which is placed on a finite level (= every digit with a finite index) is in the union of all finite trees. Therefore every path containing nodes on a finite level (= every sequence of digits at places with finite indexes) is in the union of finite trees. There is nothing remaining! > > > In that case do *not* talk about unions of sets of paths. > > > > I have a union of all finite trees. The result is an infinite tree > > (which need not be an element of the union). > > But you are talking about the union of sets of paths. What is the > relevance? An infinite tree contains all possible paths. The uninon of all finite trees is the infinite tree. Other infinite trees are not available. There is no real number available which is not represented in the union of all finte trees. Nevertheless, the representations in the union of all finite trees are countable as the countable union of finite sets. Regards, WM
From: Virgil on 8 Jan 2007 16:48 In article <45a25d13(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > > A number is not a digit string. The fact that there are no rationals > > between two reals constitutes proof that those reals are identical. > > See the reference above for an explanation of Dedekind cuts. > > > > > > What if one of those reals is a rational itself? Is there necessarily a > rational between every rational and every real, or are there irrational > reals "adjacent" to rational reals? Given any dense subset of the reals, the absence of a member of that set between two supposedly different reals, regardless of the nature of those reals, proves them equal. And in dense sets, such as the reals or rationals, there is no such thing as "adjacency".
From: Virgil on 8 Jan 2007 16:53 In article <1168271417.069985.286650(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > That L_D does not exist. > > > > Assume a line that contains any element that can > > be shown to be in the diagonal exists. > > Otherwise the diagonal could not exist, because it is, by definition, > the union of line ends in the EIT. > > > Call > > this line L_D. > > > > A: L_D is a line, therefore L_D has a largest > > element. > > > > B: L_D contains any element that can be shown to exist > > in the diagonal, therefore L_D does not have a largest > > element. > > > > Contradiction. Therefore L_D does not exist. > > Therefore the union of line ends including the end of L_D does not > exist. It is any L_D which can not exist since it would be required to have both a last member and also the successsor to its last member. Fine. There are no infinite sets. There is only potential > infinity. Every line including the diagonal The diagonal, by not having a last member, is not a line, all of which, by definition, must have last members.
From: Virgil on 8 Jan 2007 17:09 In article <1168290342.751050.317770(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > No one, except possibly WM, is claiming that an infinite binary tree is > > > > the same as the union of all binary rational, and therefore finite, > > > > trees. > > > > > > > What else could it be? > > > > It could be, and is, the union of infinitely many finite trees. > > Since the union of two trees is not a tree, whyever should the union of > > more than two trees be a tree? > > The union of two trees is a tree. Then it will have nodes which are simultaneoulsy terminal nodes and not terminal nodes. One can embed one tree into a larger tree, but that operation is not union. >That is the same as the union of two > initial segments of natural numbers which is an initial segment of > natural numbers. > > The union of the one-level tree > > 0. > / \ > 0 1 > > and the two-level tree > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > > is the two-level tree > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > > similar to the union of the segment {0,1} and the segment {0,1,2} which > is the segment {0,1,2}. That is merely the embedding of the smaller into the larger, it is not a union. > > > > > > > If it is something else: which nodes are distinguishing it from the > > > union of all trees? > > > > The fact that a union of two or more trees has more than one root node > > distinguishes that union from any single tree. > > The finite union of two or more finite trees is a finite tree. An > infinite union of finite trees is the infinite tree. The union of these two trees is not a tree at all: A D / \ / \ B C E F > > > > > > > > According to set theory, most numbers have no names, no addresses and, > > > therefore, most probably, no existence. > > > > What axiom requires that a number have either a name or an address? > > No axiom requires this but the definition of a number as an imaginated > item requires that an individual number needs an individual address to > think about it. What definition is that? I do not find any such definition in any of my math books. > > > > > > ========================================== > > > > > > >> So where are the nodes and edges which make the infinite paths longer > > > >> than any finite path? > > > > It is the absence of any terminal edge or node in an infinite path that > > makes it different from a finite path. > > In the union of all finite paths there is no final node. But in a union of trees one does not get a union of paths. > In the unuion of all last levels of the set of finite trees there is no > last level. > > > ========================================== > > > > > > > I do not know what sort of trees WM is dreaming about, but in my > > > > infinite binary trees, at each node there are infinitely many paths > > > > branching left and infinitely many others branching right. > > > > > > Is in the union of all finite trees one node through which *not* > > > infinitely many paths > > > are branching left and infinitely many others are branching right? > > > > As each node in the union is a member of a particular finite tree and > > each node in a finite tree lies in only a finite number of paths, yes. > > Each node lies in an infinite number of paths. Or which node does not? A node in a union of trees,is not a member of any tree to which it did not originally belong before the union was formed. If WM wants to embed trees so that each node of each tree is matched up with a corresponding node in each larger tree, that is quite different operation, which WM has not yet described. > > Regards, WM
From: Virgil on 8 Jan 2007 17:13
In article <1168291077.860958.62580(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: The union of the one-level tree A / \ B C and the two-level tree D / \ E F / \ / \ G H I J Is a set with ttwo disjoint trees in it. |