An uncountable countable set
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From: Virgil on 5 Jul 2006 18:21 In article <1152133409.564175.47720(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > 0.1 > > > > > 0.11 > > > > > 0.111 > > > > > ... > > > > > 0.111...1 > > > > > ... > > > > > > > > > > But in this list the number 0.111... is not contained. Hence not all > > > > > of > > > > > its digits can be identified. > > > > > > > > You have just proved that your example supports Cantor's theorem by > > > > producing a number, 0.111..., not in your own list. > > > > > > All digits of a number must be indexed by natural numbers. Otherwise > > > they cannot be identified. All digits which can be identified are > > > pesent in the list which contains all unary representations of naturlal > > > numbers. Those are not natural numbers at all in the list, but decimal, or other base, fractions. It is the digit positions in those numbers that are indexed by naturals and there is no more an end to the naturals used to index them than there is to the digits being indexed. > > > > > > That presumes that the set of naturals is finite, which in ZFC and NBG > > is false. > > It presumes only that *every* natural is finite. It presumes a last natural to be usable as an index, which presumes that the set of naturals is finite.
From: Dik T. Winter on 5 Jul 2006 18:47 In article <1152134107.848676.131670(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > David Hartley schrieb: .... > > So you acknowledge that your process doesn't give, in the limit, a > > bijection N -> Q+ ? > > Such a bijection does not exist, because there is no smallest positive > rational. There is no need of such for a bijection. Consider the set of positive numbers P and the set of negative numbers N. Clearly there is a bijection from P to N: f(x) = -x. But N has no smallest element. A bijection is not necessarily order preserving. -- 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 5 Jul 2006 18:54 In article <1152134779.638396.157500(a)b68g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > You consider aleph_0 to be a number. I think this is a problem, but I let > > it pass. > > It is not a number, but it is said to be a number. > > > > > And to > > > make sure that every digit is covered by your proof, you must find out > > > whether "if" or "else" has to be applied. > > > > Why? We know that either "if" or "else" has been applied, that is sufficient. > > The same is with my transpositions. But the order type of your set of transpositions is w * w. Moreover, different orderings of the same subset of transpositions will yield different results. In the formula I gave for the diagonal number, calculation of one digit does *not* depend on the calculation of another digit. > > > > Apparently you do not understand what Cauchy involves. > > > > > > Do you know it better than you know Cantor's writings? > > > > > > There is the positive epsilon arbitrarily small. But every epsilon > > > covers infinitely many digits of Cantors diagonal. To follow Cantor, on > > > the other hand, *every* digit has the same weight as the first one. > > > There is no limit-process at all. > > > > Apparently you do not know Cauchy well. According to Cauchy, any number > > that is written decimally has a limit, and that limit is a real number. > > And, no, I have not read much from Cantor, so I have no idea how I should > > interprete your representation here. > > But what you claim above to be Cauchy's is exactly Cantor's (and > before him Meray's) theory of irrational numbers. Not Cauchy's. He is > that guy with the epsilon. It is Cauchy's theorem that proves that the limit exists, using the epsilon argument. > > > No problem. For a given well-order to start with, every result is > > > determined. > > > > When you use finitely many overlapping sequences of "infinite permutations". > > When I stay within the domain of finite numbers. You do not. Or you are re-ordering a subset of the rationals, not all rationals. > > For finitely many "infinite permutations". Just like it is for a list > > (which is a map from N to something, where N contains finite elements > > only). > > If the transpositions are well-ordered, they are indexed by finite > numbers - up to the last one. Wrong. In the first place, there is not necessarily a last transposition, and in the second place they are not necessarily indexed by finite numbers. Consider the following (well-ordered) sequence of transpositions (where I use, as you do, index numbers): (1, 2)(3, 4)(5, 6)(7, 8)...(1, 3)(2, 4)(5, 6)(7, 8)... what is the finite index number of the transposition (1, 3)? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 5 Jul 2006 19:19 In article <1152133514.267011.143920(a)a14g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > Try to map R on the set {1, 2, 3, K} and you will see it. > > > > R surjects to {1,2,3,L} with no problem, when L is not "mueckenh"'s > > circularly defined K. > > > > When one restricts the functions one is allowed to use to functions > > which cannot exist, as "mueckenh" keeps doing, one canot do much math. > > Therefore one should not use such functions and sets to prove > fundamental theorems. One does not. To show that f:N --> P(N) cannot be a surjection on only need show that given any function that there is a set of naturals in P(N) that that f cannot have as a value, and for any given f, {x in N: x not in f(x)} is such a member of P(N) so that that f cannot be surjective onto P(N). However, for codomains other the power set of the domain, such a restriction does not always hold. Any set S, f: S --> P(S) invokes a similar restriction, but for f:R --> P(N) there is no such restriction.
From: David Hartley on 5 Jul 2006 19:20
In message <1152134296.084553.209780(a)a14g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de writes > >David Hartley schrieb: >. >> > >> >The bijection (one-to-one corespondence) between rationals and naturals >> >is preserved. This is if not a well-order so at least more then set >> >theory can supply for the natural order of Q. >> >> Your original claim was that your process would produce a bijection from >> the naturals to the rationals which still respected the ordering, which >> is now the usual one. This would of course be the contradiction you are >> looking for, but you have not offered any attempt to prove it. > >If infinity existed and if infinite processes could an be brought to an >end, then this result would unavoidably occur. But it does not because >it cannot because there is no actual infinity. Infinite processes can be brought to an end if they have a clearly defined limit. You haven't defined what you mean by the limits of the orderings or mappings produced by your process. To make "this result occur unavoidably" you need to give definitions of both limits and show that the limit of the mappings is an order-isomorphism from N to the limit of the orderings. Using the definitions I offered, the sequence of orderings does have a limit, and it is the usual ordering of the positive rationals. The limit of the associated sequence of mappings from N would be an order-isomorphism if it existed but - no surprise - it doesn't. Unless you at least try to define these limits, and to prove the required relationship between them, I see little point in continuing this discussion. -- David Hartley |