An uncountable countable set
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From: mueckenh on 23 Aug 2006 16:09 Dik T. Winter schrieb: > > It is the same with the staircase: If the total height is H, then there > > must be at least one stair of height H. > > Wrong. Think asymptote. Now I understand your idea and your error. You intermingle the height of the staircase and its least upper bound, which is not the same. As long as you cannot find a stair that has height 1 you cannot assert that the staircase had the height 1. But even if you argue that no smaller quantity can be named as height of the staircase, then I must tell you that this fact is solely due to the decreasing differences of the stairs, converging to zero. In case of the staircase of the natural numbers this is not the case. There is no asymptote. Hence your comparison of 1 and oo fails. > > > > 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. > > > > You forget that the numbers of my list leave no other outcome. > > But they do. No. Not the stairs with constant height difference. > > > > > > 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. > > > > We have infinitely many differences between natural numbers. Hence they > > sum up to a non-natural number. Contradiction. > > What is the contradiction? There is nothing in mathematics from which you > can find that "an infinite sum" of natural numbers is a natural numbers. > (And I put that in quotes because that is not really defined in > mathematics.) As long as we are in the naturals we know: Each natural is the sum of all preceding differences. Infinitely many differences require an infinite sum. If you insist that there are infinitely many naturals and if infinity is a number aleph_0, then there is also a magnitude aleph_0. Regards, WM
From: mueckenh on 23 Aug 2006 16:10 Dik T. Winter schrieb: > In article <1156149324.184708.191200(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > Now, if we complete the tree (and the list above) the number > > > of edges is still countable, but *none* of the edges terminates at 1/3. > > > You want "a last line" for completion, but there is none. > > > > I don't want a last line, nevertheless the set of all edges is > > countable. > > So 1/3 is not in your tree. 1/3 is that path which turns left, right, left, right, left, right, ... and never ceases to change its direction. Which edge do you miss? Which dge could not be enumerated? Do you believe that the set of edges of 1/3 is uncountable? Regards, WM
From: mueckenh on 23 Aug 2006 16:12 Dik T. Winter schrieb: > > Indexing is covering due to the form of the numbers of my list. > > If position n can be indexed, then 1 to n are covered. > > In repeatal mode again? That is no proof for either of my statements. > Each position (say n) of my number can be indexed, so each segment > from positions 1 through n can be covered. This does *not* mean > either that my number can not be indexed or that my number can be covered. Then prove your assertion. Show us how your number can be completely indexed while it cannot be covered. 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. Give us at least one example how you index a number without covering all the preceding numbers. Then I will accept your assertion. Otherwise I will not - and I think no one has a reason to accept it > .. > > No. Because my number can 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. Regards, WM
From: mueckenh on 23 Aug 2006 16:13 Dik T. Winter schrieb: > In article <1156149533.333542.222290(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > But according to Cantor width 1 *is* reached while height 1 is not > > > > reached, infinity in number does actually exist infinity in size does > > > > not exist. > > > > > > I do not think so. What Cantor states is that a container exsits that > > > has both width and height 1. > > > > False. According to Cantor the height is never 1, width is 1. > > 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. Regards, WM
From: mueckenh on 23 Aug 2006 16:16
Dik T. Winter schrieb: > > What is the sum of infinitely many finite numbers? What is the sum of > > all finite numbers? > > I state that it holds for infinitely many 'n'. Not that it holds for > infinite 'n' (whatever that may be). Infinitely many have infinitely many differences. If you sum them up then you get an infinite sum. > > > But I am doing so in order to show that his arguing concerns > > impredicative definitions and is inconclusive.f is the mapping, n is a > > natural number, M_f(n) is a set which contains all nongenerators, > > including n if not including n which is mapped on M. > > 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. > 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. Regards, WM |