An uncountable countable set
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From: Virgil on 19 Aug 2006 19:02 In article <ec82bq$8q0$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > The best we can do in assuming any infinite number is to assign a value > to the number of points in the unit interval, and try to use that for > proper measure, and see if it works. It does. :) > > TO Not outside the boundaries of TOmania.
From: Virgil on 19 Aug 2006 19:09 In article <1156019242.379074.274170(a)i3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > The staircase is not an independent individual, but it is nothing than > the set of stairs. Without a stair surpassing the height H the > staircase cannot surpass height H. Without a stair surpassing every > finite height the stair cannot surpass every finite height. (surpassing > every finite = beng infinite). But the length surpasses every finite > length. But there is no single stair that surpasses every height. "Mueckenh" is indulging in the quantifier dyslexia which conflates "for every x there is a y such that..." with "there is a y such that for every x..." "For every finite x in N there is a y in N such that y > x" is true. "There is a y in N such that for every finite x in N y > x" is false.
From: Virgil on 19 Aug 2006 19:13 In article <1156019385.230586.56960(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > No. This moronic nonsense would only be necessary if the complete set > of natural numbers would actually exist and have a cardinal number > larger than any natural number. Then "this moronic nonsense" is necessary in ZF and in ZFC and in NBG. When, if ever, "Mueckenh" comes up with his own set of axioms and can prove in that system that "this moronic nonsense" does not hold, he can play his silly games in his own system, but without a system, "Mueckenh" has nothing at all.
From: Dik T. Winter on 19 Aug 2006 19:17 In article <1155998763.083248.291940(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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). I will look it up when I am back at work. > > 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. You apparently do not know what the word "computable" does mean. To be precise: all algebraic numbers are computable in the mathematical sense. Also 'e' and 'pi' are computable in the mathematical sense. And the diagonal is computable for each given list. The problem with the computable numbers is that the complete list is itself not computable, and so the diagonal is not computable. The first definition of computable number is by Turing: a number is computable if there is a Turing machine that given a natural number n as input produces the first n decimal digit of the number. Later the definition has been refined a bit. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 19 Aug 2006 20:01
In article <1155999150.167954.72660(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. No. *Both* definitions are correct, there is no contradiction between them. It is your later assertion that is incorrect: "If all digit positions can be indexed, the number is in the list". It does not follow from your definition. For my number all digit positions can be indexed, but it is not in the list. From "all digit positions can be indexed" does *not* follow that the number is in the list. That is a leap you make again and again. > Either there are not all natural numbers in my list and nowhere else, > or your number cannot be indexed completely. Nothing of that. Give me an index position in my number that is not in the list. (By definition, there are none.) But all natural numbers are in your list, as are all index positions of my number. But the number itself is not in the list, because it is not a natural number. > > > 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. Indeed, that would be a contradiction. But I said "no" to your statement "that is nonsense". And I said "yes" to the second statement. > > > > > > 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. All digit positions are indeed in the list, and it is itself not in the list. Why must it be in the list itself? Consider the digit positions as bricks. I state the number I have is built of bricks, but I do not state that the number I have is a brick itelf. That is what *you* are claiming. > > > > 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. Again back to more obfuscation I see. "All positions are, by definition finite sequences." Yes, I never told otherwise. As I said, each finite initial segment of K is a finite sequence. But, as there are infinitely many natural numbers = indices, there is also an infinite sequence in K, and that is (by definition of the word infinite = not finite) not a finite sequence. > > > > 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. Why is it impossible within the context of the axiom of infinity? My definition is quite clear, I would think: K[p] = 1 for all p in N and without any other digit is granted by the axiom of infinity (the set of all naturals exists). You state it is impossible (in the context of the axiom of infinity). Again, why? > > > 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? I would think so, yes. > > but it is not a natural number. I see > > no contradiction. > > Of course. I would be highly astonished if you did. Apparently you see one. What is it? > > > 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. Yes, indeed, in mathematics anything can exist. If I say: thing T has properties P1 to Pn using axioms A1 to Am T exists *within the context of the axioms A1 to Am*, unless I can prove that there is a contradiction between the properties. (I omit the intermediate process of giving definitions.) But you have some particular meaning of exist in mind, that I have not yet properly seen. > > > 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. > > Wro |