An uncountable countable set
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From: Virgil on 19 Jun 2006 14:43 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.
From: Virgil on 19 Jun 2006 14:47 In article <1150718841.726873.223390(a)g10g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1150654604.323294.139860(a)f6g2000cwb.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > 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. > > > > But is the set K in M? Pray give a proof. > > Of course it is, by definition, for without K M would not be M. > > Regards, WM 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 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.
From: Virgil on 19 Jun 2006 15:01 In article <1150728187.749465.36710(a)g10g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1150718841.726873.223390(a)g10g2000cwb.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > In article <1150654604.323294.139860(a)f6g2000cwb.googlegroups.com> > > > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > 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. > > > > > > > > But is the set K in M? Pray give a proof. > > > > > > Of course it is, by definition, for without K M would not be M. > > > > Ah, I see. The set M is defined depending on f. Well in that case f > > is clearly not a bijection between N and M. > > That is correct. > > > This does not tell us that > > there is *no* bijection (say g) between N and M. > > 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. > > > 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. > > > Suppose f is a bijection between > > N and S, where S is the set of finite subsets of N (such a bijection > > does exist). Construct K(f) and next M(f). Clearly f is not a bijection > > between N and M(f). However it is possible to construct a bijection > > between N and M(f): > > 1. g(0) = K(f) > > 2. g(n) = f(n - 1) when n > 0. > > So also M(f) is countable. > > That is not at all the way a bijection has to be constructed between |N > and M. That is simply impossible! That you do not like it does not at all mean that it is impossible. 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. 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.
From: Virgil on 19 Jun 2006 15:07 In article <1150728692.949790.258040(a)h76g2000cwa.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. > > > > It's not exactly clear what you are talking about here. Do > > you mean that M is the set of all finite subsets of N? > > Above I spelled out clearly that M contains the set of all finite > subsets of |N and one more set K. What is unclear? > > > In > > that case, there is definitely a bijection between M and N: > > If M was the set of all finite subsets of N, that would be easy enough > to see. > > > If you let K = { k in N such that k is not an element of f(k) } > > then we can see that > > > > K = N > > Such a mapping is possible with no doubt. There remains only one > question: what number is mapped on K? That raises the question of whether such a set as M_f can exist at all. If it cannot exist, then the issue of whether a non-existent set is countable is irrelevant. So that Meucken's first duty is to prove that a function having the sort of codomain, M_f, he describes can exist at all. If ever that is established, the construction of a bijection between it and N is fairly trivial.
From: Tim Peters on 19 Jun 2006 18:53
[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. [Dik T. Winter] >> Ah, I see. The set M is defined depending on f. Well in that case f >> is clearly not a bijection between N and M. I'm not sure I'd be so generous here. K isn't a definite set before f is specified (meaning that for K to provably exist by the axiom of separation, f has to provably exist first), but to specify f K has to be a definite set first (since K must be the image under f of some natural k, f can't be proven to exist unless K can be proven to exist first). 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. [mueckenh(a)rz.fh-augsburg.de] > That is correct. >> This does not tell us that there is *no* bijection (say g) between >> N and M. > M depends on f. And when you take g, then M depends on g. That is the > essence of M. And if Dik constructs g1, while I construct a distinct g2, the definition of M frustrates both claimed bijections simultaneously? Cool ;-) > ... |