An uncountable countable set
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From: mueckenh on 18 Jun 2006 14:16 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
From: Virgil on 18 Jun 2006 15:30 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. Using the 0 origin naturals for |N, every finite subset, S, of N bijects with a unique natural number under the mapping g:M -> |N: g(S) = sum_{x in S} 2^s so that the inverse of the f function is precisely the function that Muecknh has declared not to exist. If one uses the 1 origin naturals, the function is g(S) = sum_{x in S} 2^(s-1)
From: Rupert on 18 Jun 2006 19:19 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. > You're using the notation "f" in two ways. First you're denying that a function f with certain properties exists, then you're defining M in terms of some fixed function f, which it's not clear what it is. Use a different letter for the two things, and the define the function in terms of which you're defining M. > This shows M to be uncountable. > > Regards, WM
From: mueckenh on 19 Jun 2006 03:43 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
From: mueckenh on 19 Jun 2006 03:49
Rupert schrieb: > 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. > > > > You're using the notation "f" in two ways. No. > First you're denying that a > function f with certain properties exists, then you're defining M in > terms of some fixed function f, f is not fixed by any prescription. > which it's not clear what it is. Use a > different letter for the two things, and the define the function f is not restricted by any definition. Any mapping f: |N --> M is allowed. > in > terms of which you're defining M. Let p and q be two natural numbers and let sqrt(2) = p/q. Regards, WM |