Cantor Confusion
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From: Dik T. Winter on 2 Jan 2007 20:28 In article <1167738410.654375.175860(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1167514645.819727.261070(a)h40g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > 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. Well, have a look at ZFC+V=L. Apparently a well-ordering of the reals is automatic with V=L. > > > 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 That is easy: {x} is one of those ordered sets. Here it is the first and only 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. Wrong? I don't know. But I think a quantifier problem again. I state (P is set of paths, E is set of edges, and p1 and p2 are different): (1) forall{p1 and p2 in P} thereis{e in E} e in p1 but not in p2 or every two different paths differ in at least one edge. (2) forall{e in E} thereis{p1 and p2 in P} such that e in p1 and in p2 or for every edge there are two different paths that share that edge. For some reason you conclude from this: (3) forall{p1 in P} thereis{p2 in P} forall{e in p1} e in p2. or there are different paths that share all edges. Now (3) is clearly in contradiction with (1), so I wonder how you come to this conclusion. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: William Hughes on 3 Jan 2007 03:19 mueckenh(a)rz.fh-augsburg.de wrote: > 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. This is not possible. If we assume that not all the elments of the diagonal exist then we get ancontradiction. However, this is not the contradiction I am talking about. > > > 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? Yes. A trivial contradiction which could be used to show that L_D does not exist. However, since I make no assumption about whether or not all the elements of the diagonal exist, this is not the contradiction I am talking about. > > I think it is of little value to continue this discussion. You always > implicitly assume the actual existence of all natural numbers. No. I use three properties: i: There is no last line ii: L_D contains any element, that can be shown to exists in the diagonal. iii: It is always possible to find another element that exists in the diagonal. If you use these three properties (all of which come directly from you) then you can show a contradiction whithout making any assuption about whether all the elements of the diagonal exist. The fact that if you also assume iv: Not all the elements of the diagonal exist you can get a second contradiction. is true but irrelevant. One can reach a contradiction without making assumption iv, and making assumption iv does not make this first contradiction go away. - William Hughes
From: mueckenh on 3 Jan 2007 11:58 David R Tribble schrieb: > David R Tribble schrieb: > >> Well, now I'm confused. Could you provide an example of a natural > >> number that does not exist? > > > > mueckenh wrote: > > Take the first 10^100 digits of pi (if you can - but you cannot). It is > > impossible to bring this number to existence in the whole universe. > > Can you "bring into existence" the number 1? Here it is: .. Regards, WM
From: mueckenh on 3 Jan 2007 12:09 Dik T. Winter schrieb: > In article <1167738133.535489.164800(a)n51g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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. > > You state so, contradicting the axiom of infinity. I do not contradit the axiom of infinity by free will. I am forced to do so. Otherwise I had to assume that the union of all finite binary trees is identical to the complete binary tree, as far as nodes and egdes are concerned, but that the trees are not identical as far as paths are concerned. As the paths consist of nodes and edges and of nothing else, this would not be tolerable for me. Therefore I refrain from believing in actual infinity. > > > But if you insist that > > irrational numbers exist, then you must also accept the existence of > > 1/3 as a binary series. > > It exists. > > > > > 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? > > No, it has the same number of edges and nodes, but it has more paths. Sic! I cannot accept ghosts in mathematics, but, of course, I cannot prove that they do not exist. So we are finshed with this topic. Regards, WM
From: mueckenh on 3 Jan 2007 12:27
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > Here is the EIT fort newcomers: 0.1 0.11 0.111 .... 0.11111 is the unary representation of the natural nunmber 5 L_D is the diagonal (0.)111... > > I think it is of little value to continue this discussion. You always > > implicitly assume the actual existence of all natural numbers. > > No. I use three properties: > > i: There is no last line Then also the number 1/9 = 0.111... does not exist as a decimal representation? I agree! But this implies he nonexistence of the diagonal. > > ii: L_D contains any element, that can be shown to exists > in the diagonal. Yes. > iii: It is always possible to find another element that exists > in the diagonal. That is the same with the lines. Why should the diagonal exist actually but he system of lines should not exist actually? (1/9 = 0.111...) > > If you use these three properties (all of which come directly from > you) then you can show a contradiction whithout making > any assuption about whether all the elements of the diagonal exist. > > The fact that if you also assume > > iv: Not all the elements of the diagonal exist Not all natural numbers in unary representation 0.1, 0.11, 0.111, ... exist. If all elements of the diagonal exist (which are the last digits of the unary numbers) then these unary numbers must exist too, as far as I understand existence. > > you can get a second contradiction. is true but irrelevant. One can > reach a contradiction without making assumption iv, and making > assumption > iv does not make this first contradiction go away. The only contradiction I can see is that you assume the existence of the last digits of all natural numbers (in unary representation, forming your complete diagonal) while you do not accept (or pretend not to accept) the existence of the numbers themselves. This is unacceptable in my eyes. Regards, WM |