An uncountable countable set
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From: Virgil on 22 Jun 2006 03:22 In article <1150958128.944687.314360(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1150875562.782681.289070(a)g10g2000cwb.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > > > A set is required which is the image of k if it is not the image of > > > > > k. > > > > > A barber is required who shaves himself if he does not. > > > > > > (In case f should be surjective.) > > > > > > > 1. Given a mapping f: N -> P(N), the set K(f) constructed according > > > > to > > > > Hessenberg *does* exist. > > but it is not possible to determine that K(f). > > > > > If f were a bijection it is required (by > > > > the *definition* of bijection) that K(f) (because it is an element > > > > of P(N)) is the image of some element of N, but it is not, showing > > > > that f is not a bijection. P(N) does not depend on f, > > But P(N) \ K(f) does. Map f : N --> P(N) \ K(f), and your proof > vanishes. > > > > > so there is > > > > no mapping between N and P(N) that is a bijection. > > Or there is no complete set |N and hence no complete set P(|N). > > > > > > Who told you? > > > > The proof above. > > It does not disprove a surjection but the completeness of |N and P(|N). > > > Why do you think that non-dependence on f is a special feature? The set > K does exist for every mapping f. The mapping f cannot exist if one requires, as Muecken does, that there be some n for which f(n) = {x: x not in f(x)}. That condition makes f an impossibility. > > > Map f : N --> P(N) \ K(f). That has not been proven > > > impossible. Yes it has. Since there are bijections h: P(N) \ K(f) --> P(n), any bijection f:N --> P(N) \ K(f) implies a bijection hof: N --> P(N) Which cannot occur since function hof cannot map onto K(hof).
From: Dik T. Winter on 22 Jun 2006 05:43 In article <1150957437.491685.293410(a)g10g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > > A *surjective* mapping does not exist. The same does Hessenberg. > > > > A *surjective* mapping does exist, but it is not f. There exist a > > surjective mapping g -> M(f). > > Like this one?: > > 2) 0.12324389 > 3) 0.23123123 > 4) 0.85348714 > 5) 0.11133333 > 6) 0.31415161 > .. > > 1) 0.24446... > > This is the proof, that the proof that a surjective mapping f: |N --> > [0,1] does not exist, does not exist, isn't it? This is plain nonsense. -- 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 22 Jun 2006 05:52 In article <1150958128.944687.314360(a)i40g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1150875562.782681.289070(a)g10g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > > > A set is required which is the image of k if it is not the image of k. > > > > > A barber is required who shaves himself if he does not. > > > > > > (In case f should be surjective.) > > > > > > > 1. Given a mapping f: N -> P(N), the set K(f) constructed according to > > > > Hessenberg *does* exist. > > but it is not possible to determine that K(f). Why not? > > > > If f were a bijection it is required (by > > > > the *definition* of bijection) that K(f) (because it is an element > > > > of P(N)) is the image of some element of N, but it is not, showing > > > > that f is not a bijection. P(N) does not depend on f, > > But P(N) \ K(f) does. Map f : N --> P(N) \ K(f), and your proof > vanishes. As I have shown already in a much earlier thread, the answer is *no*. > > > > so there is > > > > no mapping between N and P(N) that is a bijection. > > Or there is no complete set |N and hence no complete set P(|N). That is altogether another theory. When you talk about current theory there is a complete set N and a complete set P(N). You may of course start your own version of set theory where that is not the case. > > > Who told you? > > > > The proof above. > > It does not disprove a surjection but the completeness of |N and P(|N). You are completely wrong. How does it prove that? > Why do you think that non-dependence on f is a special feature? The set > K does exist for every mapping f. But for every mapping f there is no k > to be mapped on K. > > The set K does not an cannot exist if f is required to be a surjection. > Why do you think, this can and must happen for a countable image set M > = S(|N) U {K(f)} ? Why do you think the reason is different from that > of P(|N) ? Let's see. Let S be the set of finite subsets of N. And let us have a bijection f: N -> S (they do exist). Now obviously f is not a bijection from N to M(f), that is clear by the construction. On the other hand, we can construct a bijection from N to M(f), as before: g(0) = K(f) g(i) = f(i - 1) when i > 0, given that f is a bijection between N and S, it is easy to see that g is a bijection between N and M(f). Ah, I hear you mutter. But g is not a bijection between N and M(g). No, of course not. g is not even a mapping from N to M(g). > > > Map f : N --> P(N) \ K(f). That has not been proven > > > impossible. > > > Then map g(n+1) = f(n) and g(1) = K(f). There is no proof that this > > > cannot be a bijection. > > > > Yes. See above. For *any* mapping it is shown that it is not a bijection. > > The same is true for any M = S(|N) U {K(f)} defined by a *surjective* > mapping f. The set K then does not an cannot exist. Hence, there is no > helping g. Note again: M depends on f, a fact that you leave out every time. M in itself is not a set, but for every given f, M(f) is a set. -- 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 22 Jun 2006 07:22 Virgil schrieb: > So the resolution is that either f, K_f and M_f are self-contradictory > and do not exist at all if f is defined to be a surjection. That is right. And why do you believe that a self contradictory approach should prove anything (for instance about the surjectivity of certain mappings)? Regards, WM
From: mueckenh on 22 Jun 2006 07:26
Dik T. Winter schrieb: > In article <1150957437.491685.293410(a)g10g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > > > > > A *surjective* mapping does not exist. The same does Hessenberg. > > > > > > A *surjective* mapping does exist, but it is not f. There exist a > > > surjective mapping g -> M(f). > > > > Like this one?: > > > > 2) 0.12324389 > > 3) 0.23123123 > > 4) 0.85348714 > > 5) 0.11133333 > > 6) 0.31415161 > > .. > > > > 1) 0.24446... > > > > This is the proof, that the proof that a surjective mapping f: |N --> > > [0,1] does not exist, does not exist, isn't it? > > This is plain nonsense. Maybe. It is, however, the same approach as yours. If it is impossible to find a direct surjective mapping, then first define the set and then try to find a surjection. I can proudly declare in your words: A *surjective* mapping does exist, but it is not f. There exists a surjective mapping g -> M(f). Regards, WM |