An uncountable countable set
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From: Dik T. Winter on 11 Aug 2006 20:44 In article <1155315607.955759.264370(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > I am using pi but I know that it is not a number but only an idea. > > > False mathematics results only from the assumption that pi was a number > > > and that all of its digits could be enumerated by natural numbers. > > > > Ah, you are opposing mathematical terminology. Not the mathematical ideas. > > Why do you know it is not a number? What is your definition of "number"? > > A number must be in trichotomy with other numbers. A detailed > explanation would require too much space. Look into my paper "Physical > constraints of numbers" on my home page. Yes, just opposing terminology. > > > It is not in the set, but the number of the elements of the set is > > > aleph_0. Under this aspect aleph_0 should have some existence in the > > > set, namely as the number of elements. Of course it doesn't. Therefore > > > it cannot be at all. > > > > An idea, just like pi. > > But less consistent. Sqrt(2) is a consistent idea, and its square is > even a number. Aleph_0 is an inconsistent idea. You state so but have not proven so. > > > 2) If we have only potential infinity oo, then we cannot even consider > > > the whole set N and there is no equivalence class of N neither of > > > larger sets. > > > > Possibly. But when we assume the axiom of infinity, there is. > > But it entails infinite natural numbers which are not, hence there is > no omega. It does not entail infinite natural numbers. You may state that again and again, but that is just false. > > > |{2, 4, 6, ..., 2n}| < 2n > > > > > > lim [n --> oo] {2, 4, 6, ..., 2n} = G = set of all positive even > > > numbers > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| = aleph_0 > > > lim [n --> oo] n < aleph_0 (because n is finite) > > > > Now, again, you fail to define what you *mean* with n --> oo. > > The same is meant as in sequences and series like: > lim [n --> oo] (1/n) = 0 > n becoming arbitrarily large, running through all natural numbers, but > always remaining a finite number < aleph_0. Well, that is not a definition at all. The limit of a sequence like 1/n is defined as: lim{n -> oo} 1/n = L if for each epsilon there is an n0 such that for n > n0, |1/n - L| < epsilon. How do I apply that to the "limits" you are using above? > > > I would > > state that the first two limits exist and are both aleph-0, > > The first is only a set, not a number. Sorry, you are right. > > because we > > are talking about ordinal numbers. I would assert that the third does > > not exist, because there we are talking about natural numbers. But > > unless you provide proper definitions of those notations I can not say > > anything to it. > > I do not provide a definition of aleph_0. Cantor provided that > definition: A number larger than any natural number. I asked for a definition of the notation, not for something else. > > Nope, in that case you get a different picture. I did show you two > > pictures originally, one with terminating paths and one with non-terminating > > paths. You elected to discuss the former, you can not now change to the > > latter. > > > I did not change. All the paths are infinite. Every edge starts from a > node and ends in a node. In that case all paths are terminating paths. The number of paths is countable, and each path is the representation of a rational number for which the binary expansion terminates. 1/3 is not in the set of paths. > > > > > The number of edges is twice the number of paths. > > > > > > > > Yup, and each path is terminated. > > > > > > ? I thought you had understood: No path is terminated. There is no edge > > > without appending node and no node without appending edge. > > > > In the first picture, there is no edge without terminating node. In the > > second there is no edge with terminating node. > > In the naturals there is no even number without a following odd number > and there is no odd number without a following even number. In the tree > there is no edge without a node from which it starts and to which it > leads. > > But no edge and no node is terminating a path. Eh? > > What is 1/2 edge? > > Think about it. What is a half cake? What s a half horse? When our > ancestors tried the fractions they may have asked those questions. Do > you really have problems? Oh, I understand it in that sense. 1/2 edge either starts at a node and does not terminate at a node, or it terminates at a node and does not start at a node. Do you have that in mind? > > Why? I tell you that with a particular definition I get a number that > > is not on the list. And I can even show (without knowing the particulars > > of your list) that the number I get is not the n'th element on your list, > > because it is different in the n'th digit. No process at all. > > But the list is infinite. Therefore it is not sufficient to prove that > a finite segment of the list does not contain the diagonal number. But that is *not* what I prove. Pray what I read. -- 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 11 Aug 2006 20:53 In article <1155315830.492099.285030(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > Wrong. There are aleph_0 numbers, but all of tem are finite. > > > > Yes. But the width is infinite. > > The line number is infinite by the axiom of infinity. Wrong. The number of lines is infinite. There is no infinite line number. > > > > but there is *no* element of the list that has infinite width, because > > > > there is no last element. (If there were a last element, indeed, that > > > > element would have infinite width, but as there is none...) > > > > > > So there are *not* aleph_0 lines now? > > > > Ah, you are assuming that a list of aleph-0 lines has an aleph-0'th line? > > No, I assume that there are aleph_0 lines, where aleph_0 is a number > larger than every natural number. But if some lines are missing, then > there are less than aleph_0 lines. There are no lines missing. > > > I did so. I founf aleph_0 lines. But I found that they do not all exist > > > unless there is one line of infinite length. So there are less than > > > aleph_0 lines. > > > > You did *not*. You did find aleph-0 lines, but you did *not* find an > > aleph-0'th line. > > But an omega-th line, because there is a line omega + 1. Neither are true. > > > The lines o Cantor's list can be enmerated. We do not remain in the > > > finite there. The edges of the infinite tree can also be enumerated. > > > See the scheme above. > > > > That is proof by example. I do not see how the infinite tree can also > > be enumerated. Pray, again, explain. > > Do you know how the infinite list of Cantor can be enumerated? No, of course not. It is not the list of Cantor, but the list provided to Cantor by somebody else. I hope that the person that provided the list did know how the list was enumerated. > Do you know how the infinite set of algebraic numbers can be > enumerated? Yes. -- 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 11 Aug 2006 21:15 In article <J3uzrx.Apr(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <1155315217.378759.114420(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > Also you state that indexing is the same as covering. This is the > same as telling: > for all p there is an n such that ... > is equivalent to: > there is an n such that for all p ... > which is false. > > Back again to quantifier dyslexia. For a simple (finite) example where quantifier dyslexia tells the wrong things can be found at: <http://mathworld.wolfram.com/EfronsDice.html> let p and n be (different) dice numbers (from 1 to 4) the first statement would be: for all p there is an n such that ... the second would be: there is an n such that for all p ... where ... stands for the statement "the probability of throwing higher with die n than with die p is larger than the reverse". The first is true, the second is false. You pick one first, next I pick one, and I have a 2:1 chance of winning when we throw, regardles of the one you pick. When you look around a bit more there are many more interesting examples. There is even one where the situation is clear with a set of three dice of different colours, but where it reverses if you duplicate the set and start throwing with two dice of the same colour. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 13 Aug 2006 11:58 Virgil schrieb: > In article <1155209478.553820.256460(a)m73g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1154967767.420316.33010(a)p79g2000cwp.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Dik T. Winter schrieb: > > > > > > > > > > > No. As 0.111... has more index positions than each and every natural > > > > number in unary notation, the natural indexes are not sufficient to > > > > index 0.111... . > > > > > > > > > Just as N has more elements > > > > > than each and every indivual natural number is irrelevant. > > > > > > > > This assertion is impossible. Compare the differences of 1 between the > > > > naturals which would sum up to a natural number infinity if there were > > > > infinitely many differences possible. > > > > > > That makes no more sense > > > > and not less > > > > > than to say that the sum of infinitely > > > naturals being infinite prevents existence of infinitely many naturals. > > > > Correct. Both conclusions are identical. > > And false. An infinite sum of 1's is not infinite? Regards, WM
From: mueckenh on 13 Aug 2006 12:00
Virgil schrieb: > > > > But if there were actually infinitely many, namely aleph_0 lines, then > > > > already the first 10 % of lines were infinitely many. And 90 % of the > > > > lines were infinitely long. > > > > > > What is 10% of Aleph_0? If you start counting your lines, at what > > > point do you know that you've counted the first 10% of them? > > > > At what point do you know that you have all aleph_0 lines? > > When you have them all. That means never, because at least the last one does not exist. In fact, there are some more missing, because not more than 10^100 can be enumerated. Regards, WM |