An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 19 Jun 2006 04:47 In article <1150703016.762443.286860(a)h76g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1150654604.323294.139860(a)f6g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > 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. > > > > > > This shows M to be uncountable. > > > > > > Regards, WM > > > > If M consists exactly of all finite subsets of }N, Meuckenh is wrong. > > But it does not. The set M is uncountable while the set of all finite > subsets of |N is countable. > > Regards, WM Note that M depends on the particular f that has been chosen. We can indicate that dependence by writing M_f. Now for every f: |N --> M_f there is a bijection g:|N --> M_f. And that is enough to prove every M_f countable.
From: mueckenh on 19 Jun 2006 05:43 Virgil schrieb: > In article <1150703016.762443.286860(a)h76g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1150654604.323294.139860(a)f6g2000cwb.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > 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. > > > > > > > > This shows M to be uncountable. > > > > > > > > Regards, WM > > > > > > If M consists exactly of all finite subsets of }N, Meuckenh is wrong. > > > > But it does not. The set M is uncountable while the set of all finite > > subsets of |N is countable. > > > > Regards, WM > > 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 ? > > Now for every f: |N --> M_f there is a bijection g:|N --> M_f. > > And that is enough to prove every M_f countable. If you consider the mapping |N --> P(|N), there is the same remedy by showing that the mapping f : |N --> P(|N) \ K cannot be used to prove P(|N) uncountable. Regards, WM
From: Dik T. Winter on 19 Jun 2006 06:44 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. -- 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 19 Jun 2006 08:07 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
From: Dik T. Winter on 19 Jun 2006 08:53
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. This does not tell us that there is *no* bijection (say g) between N and M. 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. 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |