An uncountable countable set
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From: mueckenh on 19 Aug 2006 10:44 Dik T. Winter schrieb: > > > That they > > > are countable comes from other considerations. > > > > It doesn't matter where that comes from. In fact all numbers which can > > be constructed form a countable set. > > Proof? And pray use the mathematical sense of constructed. Not what you > think it should mean. There is no question that only countable sets of numbers can be constructed in the mathematical sense. And as only a countable set of such sets can be constructed in reality, the set of all constructible sets is countable by cardinal arithmetic. For more than 100 years this is an unquestioned fact. > > > (Unconstructible numbers were not at all taken into account > > when Cantor's proof was published.) > > Yes, right. Constructable numbers come from considerations in complexity > theory. Wrong again. Cantor talked about this countable set already in his correspondence with Hilbert. Regards, WM
From: mueckenh on 19 Aug 2006 10:46 Dik T. Winter schrieb: > Where can I find that quote? I have problems with reading utf-8. Works, p. 278, first paragraph (his famous article on the diagonal proof). > > > > > The diagonal proof certainly shows that the set of constructible > > > > numbers cannot be listed in its entirety. > > > > > > That is wrong (as I see it). There exists a list of constructible numbers > > > (countability proves that). But that list is not constructible, so the > > > diagonal number is not constructible. > > > > No infinite list (of independent numbers) is constructible, because no > > one can construct infinitely many different real numbers. This argument > > shows that the whole Cantor diagonal proof is void. > > > > But if there are any lists of constructible numbers, then their > > diagonal numbers are constructible too. And this construction does > > *not* show that there are uncountably many constructible numbers. This > > argument shows again that the whole Cantor diagonal proof is void. > > Well, let me be clear. Constructable numbers is used in two senses in > mathematics. The sense I meant is "computable numbers". So let me > refrase with this word: > The set of computable numbers is countable, but the diagonal is > not computable. No diagonal is computable, except in simple cases like my list, because no infinite list is computable, again with the exception of closed formulas and simple cases like my list. But that is completely irrelevant. Regards, WM
From: mueckenh on 19 Aug 2006 10:52 Dik T. Winter schrieb: > > > I do not only *claim* that 0.111... can be indexed by the natural numbers; > > > I can *prove* it, as I have done again and again. > > > > You supposed that all digits of 0.111... could be indexed, and than > > you proved that they can be indexed. > > No. I *defined* 0.111... such that all digits could be indexed. I *defined* that all digit positions which can be indexed belong to numbers of my list. The indexes are natural numbers and all natural numbers are in my list. And I showed that your number is not in my list. Hence one of our definitions must be wrong. Either there are not all natural numbers in my list and nowhere else, or your number cannot be indexed completely. > > > And that is simply nonsense. All finite digit positions are contained > > in the list, most several times, but every one at least one time. > > I say no to the first statement and yes to the second. There is no > contradiction. Again by definition? I define: It has to be considerer as a contradiction, if a digit position which is not in the list is in the list. > > > > > > Each finite segment can be > > > > > covered, and each finite index can be covered. But the total is not > > > > > finite, but contains only finite digit positions. > > > > The digit positions constitute the number. You cannot build a house of > > bricks and claim the total house is wooden. > > I do not claim such a thing. You do. All finite digit positions are contained in the list. Your number allegedly contains not more than those but it is not in the list. > > > > So, in your opinion the axiom of infinity is purest nonsense. > > > > No. It leads to nonsense in the framework of mathematics, for instance > > in contradicts the theorem that 0.111... is not a member of the > > sequence of sequences of 1's. > > It states no such thing. It gives the theorem that 0.111... is not a > member of the sequece of finite sequences of 1's. But all positions which can be indexed by natural numbers are at a finite distance from the point, because all indexes are finite. All are, by definition, finite sequences. > > > > > >You have > > > a right to opiniate that. But do *not* claim that you have found an > > > inconsistency with the axiom of infinity, because you have not. > > > > Otherwise I had to accept your handwaving argument that all list > > sequences could index a sequence which does not belong to the list. Why > > should I? > > Because there is a proof that a number K that I *define* as K[p] = 1 for > all p in N, and without any other digit is not in the list (because it > is not a natural number), but can be indexed (because all the index > positions are natural numbers)? What is the handwaving here. You define an impossible thing. > > > > Because the sum of infinitely many differences of 1 would make up an > > infinite number. > > That is not a proof. Indeed, the sum of infinitely many differences of > 1 make up an infinite number, are you sure? Indeed, even in set theory a sum of infinitely many 1's is infinite? > but it is not a natural number. I see > no contradiction. Of course. I would be highly astonished if you did. > > > Peano axioms, inductive set. If there is n, then there is n+1. But this > > does not mean that an infinite set does actually "exist". > > The axiom of infinity asserts that it does exist. But, alas, it does not assert what "to exist" means. Therefore everything can exist. > > The assumption that an infinite set had a cardinal number. The > > assumption that the limit omega would exist. > > There is no assumption, there is an axiom that asserts it. Wrong. The axiom does not say that the set had a cardinal number. Regards, WM
From: mueckenh on 19 Aug 2006 10:56 Dik T. Winter schrieb: > > Again a conclusion without proof. Use my definition of K: K[p] = 1 for > all p in N, and there are no other digits. Why should I? Use my definition of the indexable digit positions and there are other digits than those of numbers of the list. > Show that that number can not be indexed or show that that number can > be covered by a natural number, give a *proof* of either. > > > > > > But *all* digit positions can only be indexed by natural numbers. > > > > > There are no digit positions that can not be indexed by natural numbers. > > > > > > > > These positions are all in the list - by definition. > > > > 0.111... is not in the list by mathematical proof. > > > > > > Yes. But why should it be in the list? It is not an index position as it > > > is not a natural number. > > > > But you pretend it would consist of index positions which are all in > > the list. > > Not pretend, define. I define the list such that your definition is wrong. if all natural numbers do exist, my list is complete. > > > Now find out how that can be: All index positions of 0.111... are in > > the list, but not in the form of 0.111... . In which form should they > > exist there? > > What do you mean with "in the form of 0.111..."? The required digit positions are not contained in the list in form of this infinite number but only disperged over many finite numbers (which is impossible, of course). Regards, WM
From: imaginatorium on 19 Aug 2006 10:57
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > >> > Therefore there are not infinitely many difference[s] of 1 > > >> > between natural numbers. > > >> > > >> Your consequent is proven false (see below). Therefore your > > >> implication is false, too. > > > > > > You are in error. > > > > Where precisely is the error? > > The assertion that infinitely many differences of 1 can be provided by > finite natural numbers. You've been repeating this moronic nonsense for so long it's obviously hopeless, but just answer me a simple question. In your view of these "finite natural numbers", I presume 1 is included, and 2, and 3, and so on. If you were to simply count them, 1, 2, 3, 4, and so on, never stopping unless you came to an end, do you in fact think you _would_ reach an end? Any suggestions as to what this end would look like, given that it would entail a natural number such that adding 1 somehow failed to happen. Brian Chandler http://imaginatorium.org |