An uncountable countable set
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From: mueckenh on 10 Aug 2006 07:40 Dik T. Winter schrieb: > > Example: > > The third 1 of 0.111... can index the third 1 of 3 = 0.111 as well as > > the third one of 5 = 0.11111. > > It is a strange formulation to state that the third 1 can index the third 1 > of something else, because it is *not* the 1 that indexes, it is its > position number. I would rather formulate it as: the index number of the > third 1 can index the third 1 of something else. That is the correct formulation. It clearly shows the symmetry of indexing. > > > General: Instead of third, 3, and 5 you can choose n-th, n, and m > n. > > This shows the symmetry of the relation "indexing". > > But there is no symmetry as you appear to assume. You correct formulation above makes the symmetry obvious. Further: Indexing the digit number n is equivalent to covering the string up to digit number n. A finite string can never cover an infinite string. Even a finite string of an infinite set of finite strings can never cover an infinite string. > > So infinitely is not more than finitely? > > How do you conclude that from what I write? >From your assumption that infinitely many digits can be indexed and covered by finitely many digits. Don't forget: The infinite set of natural numbers contans only natural numbers, i.e., finite strings of 1's in unary representation. > > According to that the number of 1's of 0.111... is aleph_0 which is > > larger than the number of 1's of any finite number. This forbids > > complete indexibility. > > Why is that forbidden? The number of natural numbers is also aleph-0, > so it appears to me there is a perfect match possible. nevertheless, > you state it is impossible. By the axiom of infinity, the set of > naturals does exist, and by Cantor, the cardinality is aleph-0. Also, > that set does not have a last element. But there is no natural number aleph_0. Why can't you learn that the asserted infinity of the number of numbers is completely irrelevant. What counts (in the true sense of the word) is how many digits a number has. And every natural number has only finitely many digits, regardless of how many numbers there are. > > > > Not if you need *all* list numbers to index > > > > You say "not". That "not" is understandable but nevertheless it is > > wrong. Even if all list numbers are needed, none of them is infinite > > (by definition each one is finite; that is why there is no infinite > > set of finite numbers). > > Yes, you keep stating that, but the axiom of infinity asserts that there > is. Please stop your current intermingling of infinity in sizes of numbers and number of numbers. There is no ifinite size. from hat I conclude that there is no infinite set. > But if you claim that the set if finite numbers is finite, there is > a largest number. There is no largest number. The set of of natural numbers is infinite, but they don't exist all together. If they did, then there were infinite numbers, as you always seem to believe. > What is the successor of that largest number? I would > state that successor is also finite. A contradiction, so the set of > finite numbers is not finite. There is no infinite natural number. Because there is no infinite finite number - and every natural number is finite. Even the axiom of infinity does not state the contrary. (But if the axiom of infinity had to be satisfied, then a finite infinite number would be required. Compare the staircase.) > > Absolutely uninteresting how many there are. Each one is finite. > > Yes, and so what? There are nevertheless infinitely many of them. Can't you understand that it is uninteresting how many there are? Indexing is not executed by a many of numbers but each time by only one number. And this one number is always finite - each time! THERE IS NO INFINITE STRING OF 1's IN THE UNARY REPRESENTATION OF NATURAL NUMBERS. > > > Indexing all positions of 0.111... means covering all positions, i.e., > > covering the whole number 0.111... . If this were possible, then the > > list would contain at least one number 0.111... with at least aleph_0 > > positions filled with 1's. That, however, is impossible. > > Not so. For each n a finite segment of 0.111... is covered, but there is > no n that covers 0.111... completely. Therefore 0.111... cannot be indexed completely. > So we can cover each finite segment > of 0.111..., and we can index each finite position of 0.111... . But that is not sufficient to index and to cover 0.111... completely. > But there > is *no* n that indexes the last position of 0.111... (because there is none) That is no argument, because there is no last n too. > and so there is no n that covers all of 0.111... . Could we use 0.111... > to index a digit of itself (assuming it is some representation of w)? The > answer is *no*, 0.111... has no w-th digit. Fine. The correct conclusion is: 0.111... does not exist. w does not exist. w + 1 does not exist either. > > > > > That is squared insanity. Prove it. > > I would like that you prove your assertion first. But let me have a go > at my proof. Assume [p] (with p in N) to mean the p'th digit in the unary > representation of a number, and assume K to be the (non-natural) > number with a 1 in each position (i.e., for each p in N, K[p] = 1) and > no other 1's. What do I have to prove? Clearly, all digits of K can > be indexed, by the definition of K. By definition of the squared circle, the squared circle has four corners. > What remains is, I think, to prove > that K is not a natural number. Well, that is easy, for each n in N > there is a p in N such that n[p] != 1 (take p = n+1). So K is not > a natural number. Do you need more proof? > > (I remove not-symbols from the text, I have no idea where they come from): They are soft hyphens. > > Greater than every natural number is the omega (completed, defined, > > unchanging) only then, if it contains a digit position which is not > > reached by any natural number. > > That does not follow. It does not follow that a number which is larger than any natural number is larger than any natural number? Strict logic of set theory! How can a number be larger than any natural number if it is not larger than any natural number? If it is larger, then it contains something which no natural number contains. > > That is corr
From: mueckenh on 10 Aug 2006 07:42 Virgil schrieb: > What definition is that? I know of no definition which denies natural > number indexability to every digit in an endless string of digits. > > If one can index every rational number, and one can, That is unproven nonsense. It is only possible to index the few first rational numbers which you can name. Those which cannot be named (because numerator or denominator are larger than the largest natural ever named) cannot be indexed. This argument is *independent* of physical constraints. > indexing every > digit in a string is trivial by comparison. > Indexing the digit number n is equivalent to covering the string up to digit number n. A finite string can never cover an infinite string. Also an infinite set of finite strings can never cover an infinite string. You believe in stupid and easily refutable theorems. I would not be astonished if you believed in witchcraft and witches and if you wanted to burn them. Regards, WM
From: Virgil on 10 Aug 2006 14:49 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. > > > > Since the sum of more than two naturals must be a natural larger than > > any of them, there are more than any finite number of naturals. > > Yes, the set is potentially infinite. But it is wrong to assert that > all naturals existed somewhere, because "all naturals" makes not more > sense than "the last natural". It does in ZF. And ZFC. And NBG. Since "mueckenh" will not reveal his axiom system to us, nothing makes sense in it. > > > > If "mueckenh" wishes to claim that "more than any finite number" is a > > finite number, > > The idea that "more than any finite number" would be a number is wrong > too. NOT necessarily. Depends on what one means by "number". As mathematicians standardly use all sorts of different things as "numbers", well beyond the limitations of mere naturals, one need not feel limited to naturals only. If "Mueckenh" chooses to limit himself so, then he automatically excludes the rationals, the reals , the complexes, the quaternions, octonians, all the finite rings, and all sorts of other algebras. > > Regards, WM
From: Virgil on 10 Aug 2006 15:15 In article <1155210155.455852.79680(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > What definition is that? I know of no definition which denies natural > > number indexability to every digit in an endless string of digits. > > > > If one can index every rational number, and one can, > > That is unproven nonsense. It has been often proven. Given the set of rationals, Q, each rational represented by m/n, where m is an integer, n is a positive integer, m and n are coprime, with 0/1 being the representation of zero. Define f: Q -> N by if m < 0 the f(m/n) = 3^(-m)*5^n else f(m/n) = 7^m*11^n This injects Q into N. Define G:N -> Q by f(n) = n/1. This injects N into Q Then by the Schroeder-Bernstein theorem, there is a Bijection. In fact, numerous explicit bijections have been constructed. So "Mueckenh" is WRONG! AGAIN! AS USUAL!!!
From: mueckenh on 10 Aug 2006 16:35
David R Tribble schrieb: > Dik T. Winter schrieb: > >> The last line of the triangle will have equal width as the length of the > >> triangle. If we get at an infinitely long triangle, there is no last line. > > > > Mueckenh wrote: > > And there is no definite number aleph_0 of lines. > > > > 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? Regards, WM |