Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 2 Jan 2007 06:37 Virgil schrieb: > > > In the union of all finite trees, there are no infinite paths at all, > > > > No? Where do they end? You should be able to answer this question and > > to prove your answer. > > In any union of finite trees, every path is a member of some finite > tree, and it ends where the paths in that tree end. QED. Where is the ending of the union of all finite paths? Isn't it smilar to the "ending" of all finite natural numbers? > > > Hint: In the union of all finite initial segments of N there is an > > infinite initial segment, namely N (so some people claim at least). > > According to that definition there must be a member of a union which is > not a member of any set in the union. That is definitely non-standard > set theory. It is not standard that the object N = {1,2,3,...} is not in the union of all initial segments {1,2,3,...,n}? Regards, WM
From: mueckenh on 2 Jan 2007 06:42 Dik T. Winter schrieb: > In article <1167493176.710056.286330(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > At what point is 1/3 represented? > > > > In the binary tree it is not represented, but it could be represented > > by using a ternary tree. > > Now you are arguing differently from before. When I stated that 1/3 > was not in your tree, you argued that it was in your tree. Do you now > think differently? No. I have been knowing for several years already that the decimal or binary representation of 1/3 does not exist. But if you insist that irrational numbers exist, then you must also accept the existence of 1/3 as a binary series. > > > There is no representation > > of 1/3, unless it were in the infinite. But if you insist on an > > existing decimal representation of 1/3 in the infinite, then also the > > union of all finite binary trees contains such an existing binary > > representation of 1/3. > > I still do not understand what subtleties you are making. In the *complete* > binary tree 1/3 does exist. But that is *not* the union of your rational > trees. So the complete tree must have some more edges and nodes. Where are they? By the way, if you are right, then N = {1,2,3,...} is not in the union of all initial segments {1,2,3,...,n}? Perhaps, N does not at all exist? > > > My question is only the following: Do you think that the union of all > > rational binary trees is the complete infinite binary tree or not? > > It is not. But I already did answer that when I answered your question > what the distinction was between the union of rational trees and the > complete tree. So why do you ask again? I state they are distinct, > and indicate in what way they are distinct, and you ask whether I > think they are distinct? Sorry, you answered the question "What distinguishes this union of rational trees from the complete tree?" That some rational numbers are missing. If the union contains all rational numbers, there must be one of the rational trees that contains (for instance) 1/3. There is none. and you asked: Where is 1/3? At what point does it get into the union of rational trees? But you did not say which edges or nodes distinguish the complete tree from the union of all rational trees. Do you believe that there are such distinguishing edges but that one cannot name them? Or do you think that there are same edges in both trees but that some paths can form in one tree which cannot from in the other tree? Wouldn't both answers point to some matheology? > > > > Where is 1/3? At what point does it get into the union of rational > > > trees? > > > > > My question is only the following: Do you think that the union of all > > rational binary trees is the complete infinite binary tree or not? > > How many times must I answer that question? Once would be enough. One edge or one node would be enough. > > > > > That is the reason why the cardinal number of the paths in both trees > > > > cannot be different. > > > > > > That is the reason why the cardinal number of the paths in both trees > > > *is* different. > > > > > With a different number and configuration of nodes (or edges) or with > > exactly the same? > > You lost me here. We were talking about the union of finite trees and > the complete tree. And my question is: Do these two trees, namely the complete tree and the union of all rational trees differ such that the one has edges or nodes which are missing in the other? > You are wrong. sqrt(2) has a pretty good representation: "sqrt(2)". That is a name. Name ist Schall und Rauch. Regards, WM
From: mueckenh on 2 Jan 2007 06:46 Dik T. Winter schrieb: > In article <1167514645.819727.261070(a)h40g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > Why don't you simply give the formula for the definable well ordering? > > > > The formula could be evaluated, and, step by step, we would get a list > > > > of all reals. > > > > > > Wrong. It would not give a list. It would give a well-ordering. Do you > > > not know the difference? > > > > Every non empty set of a well-ordering has to have a first element. If > > all reals are well-ordered, then all reals are first elements in some > > sets. > > What is the relevance? If I order the naturals as: > {1, 3, 5, 7, ..., 2, 4, 6, 8, ...} > I have a well-ordering but not a list. A well-ordering is *not* a list. It is not a list in the technical meaning of "list", but it is a linear ordering. And aleph_0 linear orderings result in cardinality aleph_0. The well-ordering of the reals must contain aleph_1 omega numbers. Therefore I would be interested how to state that well-ordering. > > > > And I doubt whether you could apply the formula > > > one by one to each real. > > > > I doubt that too, in fact it is impossible in ZFC. Therefore, there is > > no well-ordering of all the reals. > > You keep to state that, without proof. For a well-ordering it is not > necessary to be able to state the elements one by one. As your above example shows, one must be able to find for every x the set where x is the first element > > > > > Because non-distinguishable sets are not different sets. You just > > > > proved that there is no element of a path which distinguished it from > > > > every other path. > > > > > Right. But that does *not* mean that it can not be distinguished from > > > every other path. > > > > If every edge of a path is shared by another path, how would you > > distinguish them? > > What is "them" here? Give me two paths and I can give you an edge that > distinguishes them (or at least an algorithm that shows it). "Them" is two paths. If one path has no edge of itself, then there is at least one other path which shares all edges with the former. Regards, WM
From: mueckenh on 2 Jan 2007 06:52 cbrown(a)cbrownsystems.com schrieb: > "3/2 < 2" and "|N| < |R|" have /completely different meanings/. The > first is a linear order applied to a field. The second is an ordering > based on surjections / injections. > Both are based upon linear orderings. > > Sums of rational numbers are nothing else but sums > > of integers, given in the unit of the greatest denominator. In case of > > limits this kind of arithmetic is extended, but the basis is > > calculating with natural numbers which are to be compared by mappings, > > i.e., the arithmetic of ordinal numbers. > > 3 is not identical to 6. 7 is not identical to 14. And yet we say 3/7 > is identical to 6/14. In fact, there are at least as many names for > the number "3/7" as there are natural numbers. This is the same as the > number of names for the number "2/9"; which is to say there is a > bijection between the names for "3/7" and the names of "2/9". > > Your thoughts? Sorry, I have not the time to develop here the arithmetic of rational and irrational numbers. But they all are based on counting of natural numbers, i.e., ordinal numbers. Regards, WM
From: mueckenh on 2 Jan 2007 06:56
William Hughes schrieb: > > > > Why? How should we know which can possibly exist? > > > > > > L_D must contain the diagonal. > > > Thus, L_D must contain any element that > > > can be shown to exist in the diagonal. > > > > It does, after the existence has been shown. > > However, before existence has been shown it does > not. Correct. Otherwise the actual existence of all natural numbers would be implied. > > If L_D contains the diagonal then it contains > any element that can be shown to be in the diagonal Yes. L_D contains any element, that can be shown in the diagonal. > whether or not existence has already been shown. > It must contain the element before existence has > been shown. How should that be possible? With L_D containing all these elements, their exisence would be deternmined. > We cannot know about all the elements that > can be shown to exist. Correct. > However, a single > example suffices. > > There is an element of the diagonal, > that has not yet been shown to exist, > but which can be shown to exist which is not an element > of L_D. Therefore, L_D does not contain the diagonal. L_D contains any element that can exist, from the first to the last one. L_D cannot contain an element which does not yet exist, because the elements (digits) of L_D are supposed to exist, but they cannot exist, if they don't exist. Isn't that a triviality? I think it is of little value to continue this discussion. You always implicitly assume the actual existence of all natural numbers. This idea bears a contradiction. But this contradiction becomes more obvious when considering the binary tree, which, according to some correspondents here, is different from the union of all finite trees. Regards, WM |