An uncountable countable set
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From: mueckenh on 26 Aug 2006 05:38 David R Tribble schrieb: > Logic > dictates that every set must have a cardinality, either zero, a > finite cardinality, or an infinite cardinality. How could the logicians live happily during the 2500 years before Cantor? > > I have never said I think there is a largest natural. I have said that > > some of your assumptions lead to that conclusion. > > You assume that the axioms of set theory lead to that conclusion, > but you've never proved it. No number of the form 0.1, 0.11, 0.111, ... can index the number 0.111... without simultaneously covering its smaller digit positions. That is proof enough to show that from the impossibility of complete covering the impossibility of complete indexing follows. All you can do is to deny that a proof is a proof. Regards, WM
From: mueckenh on 26 Aug 2006 05:45 Dik T. Winter schrieb: > And similar holds for the width. So either you state that nor the height, > nor the width are 1, or you state that both are 1. That depends entirely > on your viewpoint. But as the height and width of my staircase at step n > is 1 - 1/2^n, I would say that the height and width of the completed > staircase is the limit of that number. Apparently you disagree. > > > There is no asymptote. Hence your > > comparison of 1 and oo fails. > > Yes, I wanted to simplify the picture to show that what holds for the > height also holds for the width. > And that is just the deviation from the natural numbers. Here the 1 in width aleph_0 is asserted to exist for the set of natural numbers *without* a number omega. What TO and I are arguing is that either none of them do exist or that both exist. So aleph_0 is the cardinality of the set including the *number* omega. Less numbers are not sufficient to obtain aleph_0. Your example shows very clear that either both LUBs have to be included - or none. In my opinion we choose none. But whatever you choose, you cannot avoid to see the symmetry, can you? > > > > 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. > > Wrong. In mathematics the concept of "infinite sum" is not defined, But if there are infinitely many naturals then the infinite sum is necessary. >as I > have told you already *many* times. We can talk losely and state that > the infinite sum of all those numbers is indeed aleph-0, but note that > that is talking losely. All infinite sums as in sum{n = 1 .. oo} > used in mathematics are defined by limits. You told me that set theory does not involve limits: "As set theory does not know about limits, I fail to see that". > > > 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. > > Makes no sense, again. What do you mean with the term "magnitude"? Just the same a size. > The only connection I know of is where it is used in connection > with real or complex numbers. And it can also denote some norm > (in general Euclidean norm) in vector spaces. And as the finite > cardinals equate to the non-negative integers, that equate to their > magnitude, you might consider a cardinal number to represent its own > magnitude. Do you mean that? Exactly. And in addition, as long as the numbers are finite, ordinal and cardinal is the same size. That is just why the cardinal cannot be infinite as long as the ordinal is finite (as expressed by "the infinite number of finite numbers"). Regards, WM
From: mueckenh on 26 Aug 2006 05:47 Dik T. Winter schrieb: > > 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? > > As in your tree, by definition of the tree, each edge terminates at a node, > what is the number of the edge that terminates at 1/3? What is the number of > the edge that terminates at 1/5? I don't understand these questions. 1/3 and 1/5 are infinite paths consisting of infinitely many edges. > As the set of all edges is countable, you > should be able to state that. The problem with your counting method of > edges (count first level 1, next count level 2, etc.) you will already > have exhausted all natural numbers by counting all edges that are at the > end of finite paths. There are no finite paths and there are no ends at all in my tree. If you think that 0.1 would be represented by a finite path, what should then be represented by the path 0.1000...? By your arguing you would never be able to count the digits of 1/3 because "you will already have exhausted all natural numbers by counting all edges that are at the end of finite" sequences. > > > Do you believe that the set of edges of 1/3 is uncountable? > > That is not the set of all edges in the tree. No. But would you say that the set of all edges of 1/3 and of 1/5 and of 1/7 were uncountable? How many paths are required to obtain an uncountable set of edges, according to your opinion? Regards, WM
From: mueckenh on 26 Aug 2006 12:29 Dik T. Winter schrieb: > In article <1156364007.547996.270100(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > 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). Or direct: aleph_0 is the number of all *finite* numbers. Regards, WM
From: mueckenh on 26 Aug 2006 12:31
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. > 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. There is none! This is simply proved by the numbers of my list 0.1 0.11 0.111 ..... > > > 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. > > > 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. > > > > 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 ??? 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. Regards, WM |