An uncountable countable set
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From: Dik T. Winter on 27 Aug 2006 20:52 In article <1156609759.075864.266590(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1156364007.547996.270100(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > > I do not think so. With blocks of height 1/2^n and width 1 - 2^n, > > > > after k steps the total height is 1 - 2^k and the total width > > > > 1 - 2^k. When we complete we get the limiting case. But there is > > > > still no block at either height 1 or with width 1. > > > > > > > > And there is no smaller container that contains the complete stair. > > > > > > According to Cantor: The infinite set of finite numbers. aleph_0 is > > > actually infinite, no natural number is actually infinite. The > > > staircase has width [0, 1] and height [0, 1). It is a difference. > > > > I do not see a proof in your statements. The staircase is within > > [0, 1) and [0, 1) after completion. How is that in contradiction with > > what Cantor states? > > Cantor states (analogously) that *all stairs exist*, that the width of > all ot them is L, but that none of them has hight L. Width [0, 1] and > height [0, 1). A quote please. But indeed, the width of all of them is L, but there is none of them with width L. And the heighth of all of them is L, but there is none of them with heighth L. So width [0,1] and heighth [0,1]. That is, if you define both as the smallest box containing the completed stair. If you define both as the largest numbers that can be obtained, you get width [0,1) and heighth [0,1). -- 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 27 Aug 2006 21:12 In article <1156609912.036360.51250(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > Again, definition: K is the number with K[p] = 1 for all p in N and no > > other digits. > > Do you agree that is a valid definition? If not, why not? > > Because there is no number which can be completely indexed while it > cannot be covered. That makes no sense. A definition is not invalid if there is no number that can satisfy the definition. And in that definition I neither us the term "completely indexed", not "covered". > > K can be completely indexed because each digit position is a natural > > number. > > Do you agree with this? If not, why not? > > Because there is no number which can be completely indexed while it > cannot be covered. > > > K can not be covered because there is no natural number n such that all > > digit positions are less than or equal to n. > > Do you agree with this? If not, why not? > > Because there is no number which can be completely indexed while it > cannot be covered. This all is a prime example of circular reasoning. I show you a number that can be completely indexed while it can not be covered and ask you what parts are wrong, and the only thing you state is that such numbers do not exist. This all reminds me strongly with a discussion I had with either you or Eckard Blumschein where I showed a step by step reasoning for the uncountability of the powerset of the naturals, and where it was agreed that each step was factually correct, but the conclusion was wrong. The conclusion was the last step. > > > > > No, don't tell us that this was > > > true or that you defined that, but show us *how* you manage this trick > > > which, in my eyes, is impossible. > > > > With the axiom of infinity it is dead easy, see above. And I thought we > > were arguing with the axiom of infinity in mind. > > Yes, we do, but also with some fundamental understanding of the fact > that unary numbers are finite sequences of 1's which cannot index a > digit without covering its precursors. You cannot sacrifice the most > obvious and simplest truth only to safe the consistency of the axioms. But never did I negate that. And I always state that. > > > Give us at least one example how you > > > index a number without covering all the preceding numbers. > > > > I have no idea what you are stating here. > > You assert that all digits of 0.111... can be indexed by unary numbers > of my list without being covered by at least one of them. Give us one > example where a number of the list indexes a digit position without > covering all smaller ones. If you cannot give one finite example, then > your assertion is wrong for finite digit positions. And as there are > only finite digit positions, even if there were infinitely many of > them, then your assertion is definitely wrong. But your assertion is trye for each finite digit position. Pray reread what I wrote above. And pray show what of my assertions are wrong (and not by stating that no number can be indexed and not be covered). > > > > It is your lack of a proper proof that if each digit of a number can > > > > be indexed that number can be covered. And such a proof does not > > > > exist. > > > > > > I do not see how I could avoid my conclusion. But if you are so sure > > > then give us at least one example how you completely index a number > > > without covering all the preceding numbers. > > > > I have done so many times, and am doing it here again. > > You gave an example how a number of the form 0,111...1 with n digits > indexes the n-th digit but does not cover all digits with m =<n ??? No, because that does not exist. > Your only problem is the "infinite number of finite numbers". All the > laws of finite numbers are valid for finite numbers. Hence, if you > cannot show that your statement (indexing all but not covering all) is > correct and possibly satisfied, then you must give an example at a > finite digit position, because there are no other. You think so. I think you have a problem with "infinite number of finite numbers". Within the set of naturals there is no largest natural. -- 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 27 Aug 2006 21:22 In article <1156610108.784198.90920(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1156364184.155913.12090(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: .... > > Well, if you give me a proper definition of the sum of infinitely many > > numbers, I may have some idea about your meaning. In mathematics such > > a thing is not defined. Your statement above is not a definition. > > If there are infinitely many numbers and, hence, infinitely many > differences, then there is a definition of an infinite sum. By what? Fiat? I ask for a definition, and I note that you are unwilling or unable to provide one. > > > > Yes. Such triples do not exist. And that precisely shows why > > > > Hessenberg's proof was right. > > > > > > That it is false. An argument which makes use of the last digit of pi > > > is void, because the last digit of pi does not exist. > > > > Again avoiding the issue by going on with completely different things. > > No. Only showing by an example what you desire or believe. Avoiding te issues by going on with completely different things. Are you not able to keep to the issue involved? > > > > If there is a surjective mapping f from N to P(N) it is > > > > a requirement (of surjectivity) that such a triple *does* exist. > > > > > > I gave an example that this set cannot exist, independent of the > > > surjectivity, independent of the cardinalities of the sets involved in > > > the mapping. > > > > But it is trivial that such a triple can not exist. > > Of course. It is as trivial as the fact that the last digit of pi does > not exist. Therefore I used this example. Perhaps, I do not see a relation between the two. > > But if there is > > a surjective mapping from N to P(N), such a triple *must* exist. > > No. If a number is indexed, then it is covered (by my list numbers). Again going off at a tangent. So you are actually stating that there are surjective mappings from A to B such that an element of B is not in the image of the mapping? > You deny this. So you can believe also other absurd ideas like this: > The set M(f) may exist, but not defined by M(f) !?!? Apparently you believe many absurd things. i state that M(f) is defined by f. > I did not say that a surjective mapping would exist. I only deny that > this proof was valid. By stating that in a sujective mapping from A to B there can be elements of B that are not in the image of A. -- 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 27 Aug 2006 21:49 In article <1156689126.335154.133150(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > The first part of his first proof shows that a complete ordered field > > > > has cardinality larger than the natural numbers. In his proof he did > > > > not rely an any properties of the reals other than that they form a > > > > complete ordered field (he uses reals to exemplify). .... > > Note what I wrote: "in his proof he did not rely on any properties of > > the reals other than that they form a complete ordered field". What > > other properties of the reals did he use? > > He had no field and he used no filed. What is a field without the > axioms of the field? Nothing. Cantor disliked axioms if he did not hate > them. His only concern were numbers, numbers, numbers and their > *reality*. No fields and no axioms. Perhaps. I thought we were discussing current set theory. And not what in some time long ago was et theory. He may not have liked them, but in current terminology "the first proof shows that a complete ordered field has cardinality larger than the natural numbers". > > > He wrote that it was a simpler proof for his first theorem, i.e. the > > > theorem that the reals are uncountable. > > > > And he did *not* write that. > > Aus dem in =A7 2 Bewiesenen folgt n=E4mlich ohne weiteres, da=DF Quoting again only in part. I will quote the translation I gave, with annotation of the complete paragraph, not only the part you like to quote: In the article, titled: "?ber eine Eigenschaft des Ombegriffs aller reellen algebraischen Zahlen (Journ. Math. Bd. 77, S. 258) [hier S. 115], can for the first time be found a proof of the theorem that there are infinite sets that are not in bijection with the set of natural numbers 1, 2, 3, ..., v, ..., or, as I say in general, do not have the same cardinality as the natural numbers 1, 2, 3, ..., v, ... . So apparently he is thinking that his first article proves the theorem that there are sets that are not in bijection with the natural numbers. Which is true. From what has been proven in section 2 follows that e.g. the set of real numbers in an arbitrary interval can not be put in a sequence w_1, w_2, w_3, ..., w_v, ... . So he is now thinking that that proof was just an example for such a set. Which it is. The "beispielsweise" is telling. It is however possible to construct a much simpler proof for that theorem that is independent from the observation of irrational numbers. And the "that theorem" (in German "jenem Satze") can, in my opinion only refer to the theorem mentioned in the first sentence in this paragraph. And I think this not only holds for this translation, but also for the original. -- 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 27 Aug 2006 21:58
In article <44f1b114(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > Dik T. Winter wrote: > > In article <44ef36ae$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: .... > > > Such that one can specify which finite number of iterations will get > > > one within a specific finite range of accuracy, gven a specific > > > method of approximation? It's a limit concept, really, yes? > > > > Computer scientist you are? How wrong you are. The very first definition > > is that a number is computable if there is a Turing machine so you can give > > it a specific integer n and it will calculate all digits from the first > > to the n-th. There is no limit concept involved at all. All algebraic > > numbers are computable, as are a host of non-algebraic numbers (e, pi). > > Dik, you didn't even try to think about that one. Sorry. It was a lame > answer. Oh. > When you say you can calculate to the nth digit, But you stated "that you can specify which finite number of iterations" etc. That is something different. A number is computable when there is a Turing machine such that, when given an arbitrary number n, it will calculate the 1-st through n-th digit. There is nothing about a specification of a finite number of iterations. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |