An uncountable countable set
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From: mueckenh on 22 Jun 2006 02:19 Jens Kruse Andersen schrieb: > By definition, an infinite set M is uncountable if and only if: > There is no bijective mapping f from N to M. Correct. What do you think, why this is so? My point is only to show that the impossibility of a mapping involving Hessenberg's trick does not prove anything about cardinalities of sets involved. It shows that the sets |N and P(|N) do not exist. > > Note that M is given first, and *then* it is said that no bijective f exists > for that M, with no other condition on f than it has to be bijective. Oh, I see. Let's apply this new insight: 2) 0.12324389 3) 0.23123123 4) 0.85348714 5) 0.11133333 6) 0.31415161 ... 1) 0.24446... So this is the proof, that the proof that a surjective mapping f: |N --> [0,1] does not exist, does not exist, isn't it? > > In the well-known proof that P(N) is uncountable, P(N) is a fixed given set > before any function is even mentioned. That is the question. > The proof then shows that no matter > which function f is chosen, f will not be a bijection from N to P(N). And my proof shows that no matter what function f is chosen, there will not be a bijection from N to M. You will have obtained from my sketch of Cantor's diagonal proof above that it is *not* allowed to switch the mapping after the set has been fixed. Regards, WM
From: mueckenh on 22 Jun 2006 02:23 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? Regards, WM
From: mueckenh on 22 Jun 2006 02:35 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. 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) ? > > > 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. Regards, WM
From: Virgil on 22 Jun 2006 03:08 In article <1150957169.123006.130050(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Jens Kruse Andersen schrieb: > > > By definition, an infinite set M is uncountable if and only if: > > There is no bijective mapping f from N to M. > > Correct. What do you think, why this is so? My point is only to show > that the impossibility of a mapping involving Hessenberg's trick does > not prove anything about cardinalities of sets involved. It shows that > the sets |N and P(|N) do not exist. No more than it proves Muecken does not exist. > > > > > Note that M is given first, and *then* it is said that no bijective f exists > > for that M, with no other condition on f than it has to be bijective. > > > In the well-known proof that P(N) is uncountable, P(N) is a fixed given set > > before any function is even mentioned. > > That is the question. > > > The proof then shows that no matter > > which function f is chosen, f will not be a bijection from N to P(N). > > And my proof shows that no matter what function f is chosen, there will > not be a bijection from N to M. You will have obtained from my sketch > of Cantor's diagonal proof above that it is *not* allowed to switch the > mapping after the set has been fixed. Actually in Cantor's proof, one is allowed to construct as many functions as one likes between the fixed set N and the fixed set P(N), as the general anti-diagonal rule works equally well on any of them. Muecken insists on a function and codomain which can only exist if the function is not a bijection, but which, in that case allow a diferent fuction which is a bijection. So that either neither of Muecken's f and M t exist at all and the issue of bijection of N with a non-existent set is irrelevant. Or if f and M do exist, it is because f is not required to be a surjection, in particular K = {x in N: x not in f(x)} is not a value of f, and then f, K and M all exist, and there are plenty of bijections from N to M. > > Regards, WM
From: Virgil on 22 Jun 2006 03:09
In article <1150957437.491685.293410(a)g10g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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? No. |