An uncountable countable set
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From: Virgil on 20 Jun 2006 13:47 In article <1150796061.943937.27660(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > > [mueckenh(a)rz.fh-augsburg.de] > > >>> An uncountable countable set > > >>> > > >>> There is no bijective mapping f : |N --> M, > > >>> where M contains the set of all finite subsets of |N > > >>> and, in addition, the set K = {k e |N : k /e f(k)} of all natural > > >>> numbers k which are mapped on subsets not containing k. > Of course f does not exist. But the reason has nothing at all to do > with cardinalities. This error, however, is the foundation of > Hessenberg's proof of Cantor's theorem. f does not exist if it is required to be a bijection, but in that case neither does K(f) of M(f) exist, so there no such set as M(f). On the other hand, if f is not required to have K(f) in its image set, then there is no problem with constructing any number of such functions, and for each of them, there is a bijection (though not f) between N and M(f). > > 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. > > > > IOW, I doubt this argument can be made in ZFC, just as it's not possible > > under ZFC to prove that > > > > F = {k e |N : k e F} > > or > > G = {k e |N : k /e G} > > > > exist. > > On the other hand, why should any mapping not exist? If you claim that > a surjective mapping |N --> P(|N) must contain the set K in any case, > then there is no arguing why my mapping should not exist. The problem only arises when one insists that there is some n in N such that f(n) = K(f). If one does not make that requirement, then there is no problem with the existence of K(f) or with functions f:N --> M(f).
From: Virgil on 20 Jun 2006 14:02 In article <1150796356.837197.53290(a)g10g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1150782120.872330.199120(a)c74g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Daryl McCullough schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de says... > > > > > > > > >> >> > There is no bijective mapping f : |N --> M, > > > > >> >> > where M contains the set of all finite subsets of |N > > > > >> >> > and, in addition, the set K = {k e |N : k /e f(k)} of all > > > > >> >> > natural > > > > >> >> > numbers k which are mapped on subsets not containing k. > > > > > > > > >Above I spelled out clearly that M contains the set of all finite > > > > >subsets of |N and one more set K. What is unclear? > > > > > > > > Okay, so M is not a fixed set, but is a function of f. So to make > > > > this precise, let's put in the functional dependence: > > > > > > > > K(f) = { k in N | k is not an element of f(k) } > > > > M(f) = { A in P(N) | A is finite, or A = K(f) } > > > > > > > > where P(N) means the set of all subsets of N. > > > > > > > > Then what you are saying is that > > > > > > > > forall f: N -> P(N), > > > > f is not a bijection between N and M(f) > > > > > > > > That's true. But that doesn't mean that M(f) is uncountable. > > > > > > If it is proven impossible to set up a mapping between M and |N, then M > > > is uncountable. > > > > But you have not proven it impossible, you merely argue that ONE > > function between N and M(f) is not a bijection. > > > > > > > > WE have some f: N --> P(N), which we know is not a surjection since > > there is no n in N for which one can have f(n) = K(f). > > > > But that does no mean that there is no surjection between M(f) and N. > > > > In fact, given a function f for which K(f) exists, it is not difficult > > to build a bijection between M(f) and N as follows: > > > > For each S in M(F) define g: M(f) --> N by: > > If K is finite then > > if 0 e N > > then g(S) = sum_{s e S} 2^s > > else g(S) = sum_{s e S} 2^(s-1) > > endif > > else (K being infinite) > > g(K) = the first natural > > and > > if 0 e N > > then g(S) = sum_{s e S} 2^s+1 > > else g(S) = sum_{s e S} 2^(s-1)+1 > > endif > > endif > > > > In all these cases, g will be a bijection from M(f) to N. > > > > > > > > > > > > > We can prove > > > > > > > > forall f: N -> P(N), exists g: N -> P(N) > > > > g is a bijection between N and M(f) > > > > > > No, it is not, because M(f) is incomplete without K. But K does not > > > exist. > > > > But you have not proven it never exists, and for some functions, f, it > > Does exist. > > > > For example, let f(n) = {} for all n, then > > K = {k in N : not k e f(k) } = N. > > > > So that for SOME functions f, f, K(f) exists. > > > > The reason that Meucken's argument fails here is that there is no > > necessity for the function f to be a surjection onto M(f). > > > > The desired bijection can be established by an entirely different > > function between M(f) and N, as shown above. > > Only if K(f) would exist. If one claims that such an f is a bijection, then it cannot exist at all, but neither can K_f or M_f exist, so the problem of whether such a nonexistent set as M_f is countable is irrelevant. If one does not require f to be bijective, in particular if one does not require that there be any n in N for which f(n) = K_f, then K_f anf M_f may quite easily exist. And M_f is quite easily shown to be bijectable with N when it does exist. > > > The reason for its non-existence is not lacking number of > > > elements of |N but an impredicable definition. > > > > Wrong, as shown by a specific example above. > > Your example ist wrong because K(f) does not exist. There are two cases to consider: (1) If one requires that there be some n in N for which f(n) = K_f, then the function f does not exist, but then neither do K_f and M_f, so there is no problem as there is no such thing as M_f. (2) If f:N --> P(N) is such that there is no n in N for which f(n) = K_f, then again there is no problem, since while K_f and M_f may exist, it is easily shown that when they do, M_f and N biject. > > Regards WM
From: mueckenh on 20 Jun 2006 16:38 Virgil schrieb: > In article <1150710209.401911.130270(a)i40g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > Note that M depends on the particular f that has been chosen. > > > We can indicate that dependence by writing M_f. > > > > Oh, indeed? What is the number k mapped under f on the set K = {k e |N > > : k /e f(k)} of this M_f ? > > As soon as you tell us that M_f exists at all, YOU are telling us that > there must be such a set, K. The only other option is that there is no > such set K and no such set M_f and no such function f at all, in which > case the uncountable-countable-set becomes a non-existent > uncountable-countable-set. Of course K exists for every mapping f (if |N exists). But the set {f, k, K} is a paradox set. It is not suitable to show that a bijection |N --> {all finite subsets of P(|N) plus one further set} does not exist. Why do you believe it could be capable of showing that |N --> P(|N) does not exist? Regards, WM
From: mueckenh on 20 Jun 2006 16:41 Virgil schrieb: > > Give us an example of such an f, K_f and M_f, to show that such an f, > K_f anf M_f can actually exist. If the set {f, k, K} could actually exist, then M would be countable. But it is not, isn't it? > > If they can exist at all, it is easy enough to show that there is a > bijection beween |N and M_f, and if they can't then there is no M_f to > worry about. In order to define K_f you first have map all natural numbers n of |N on the finite subsets of |N. Then consider K_f. But there is no element k of |N remaining, which could be mapped on K. Hence the set M is uncountable. It is about the same as Cantors diagonal proof. The diagonal number depends on all numbers of his list. Moreover, after constructing the diagonal number, it turns out to belong to a countable set. My M remains uncountable in any instance. Regards, WM
From: mueckenh on 20 Jun 2006 16:44
Virgil schrieb: > > M depends on f. And when you take g, then M depends on g. That is the > > essence of M. > > Wrong! Given any two large sets are many functions for one to the other > and equality of cardinality depends on whether any one of them is a > bijection. Given two sets to compare, you cannot restrict things to only > one allowable function between them but must allow all possible > functions between them to be considered. I allow for all functions, because I did not introduce any restrictions. > > > > > In order to show that > > > there is no bijection between N and M you are not allowed to change M > > > for each and every attempted mapping. > > > > I do not change anything. I define K by exactly that mapping which is > > assumed. > > Then there are other functions between N and M_f, one of which is a > bijection. Could you prove that assertion please? > There are two issues here. The first is whether a function of the sort > Meucken tries to create can exist at all. He has given no example of > such a function, nor proof of existence. Of course there is no proof of existence. If it was, then M was countable. > > However if we allow the assumption that such a function f exists, it is > not hard to show that there are bijections between |N and M_f, though f > need not be one of them. There are many bijections between {numbers of Cantor's list plus diagonal number}. Don't you know that? Here is one: Enumerate list number n by n+1 and enumerate the diagonal number by 1. Regards, WM |