An uncountable countable set
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From: mueckenh on 21 Jun 2006 03:44 Dik T. Winter schrieb: > In article <1150835899.493068.116800(a)g10g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > 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. > > You do not show that. You show only that given a mapping f from N to the > set of finite subsets of N, the mapping f --> {all finite subsets of N > plus one additional subset conditioned by f} does not exist. A *surjective* mapping does not exist. The same does Hessenberg. > This does > *not* prove that there is no bijection between N and {all finite subsets > of N plus one additional subset conditioned by f} does not exist. The same is with Hessenberg's. > > > Why do you believe it could be capable of showing that |N --> P(|N) > > does not exist? > > Because P(N) does not depend on the mapping used. But Hessenberg's K(f) does depend on f. It is moving with f. And it can be determined for any predicable definition of f. It cannot be determined for the assumed surjection with its impredicable definition. > If you give as target > the set of all finite subsets of N plus one additional subset, I can > construct a bijection between N and that set. But your target is a > moving target. Like that K(f) of Hessenberg's impredicable definition. Regards, WM
From: mueckenh on 21 Jun 2006 03:55 Rupert schrieb: > Are you familiar with existential quantifiers and universal > quantifiers? You are making a statement of the form "There does not > exist an f such that...", or, alternatively "For all f it is not the > case that..." Then later on you use f to refer to a specific function. > You should use different letters on these two occasions. No. "For all f it is not the case" means the same as "there does not exist any f for which it is the case". A f (f =/= surj) <==> ~E f (f = sur). And I let f be any function you like. > > "Let f be an arbitrary function. There does not exist a bijective > mapping g:N->M, where M is the set consisting of all the finite subsets > of N together with the set K={k e N: k /e f(k)}." That is what I said. > > This is false. There always will exist such a bijection g. It will of > course be different to f. The same is true is you take Hessenberg's mapping of f:N --> P(N). It is not surjective because of K(f). But if you define g(n+1) = f(n) and g(1) = K(f), the proof is no longer a proof. Regards, WM
From: Rupert on 21 Jun 2006 04:02 Virgil wrote: > In article <1150870468.791288.14540(a)p79g2000cwp.googlegroups.com>, > "Rupert" <rupertmccallum(a)yahoo.com> wrote: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Rupert schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > 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. > > > > > > > > Yep. In one of the occurrences it occurs preceded by a universal > > > > quantifier, in the other it occurs as a constant symbol. > > > > > > Could you say what you mean? > > > > > > > Are you familiar with existential quantifiers and universal > > quantifiers? You are making a statement of the form "There does not > > exist an f such that...", or, alternatively "For all f it is not the > > case that..." Then later on you use f to refer to a specific function. > > You should use different letters on these two occasions. > > > > > > > > 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. > > > > > > > > > > > > > It doesn't make sense to talk about the set K={k e |N: k /e f(k)} unles > > > > you've specified what f is. > > > > > > f is not fixed by any specification. Therefore my arguing holds for any > > > f. K is that subset of |N which contains all natural numbers which are > > > not mapped under f on sets containing themselves. > > > > Okay. So let me have a shot in translating what you've said into > > something coherent. > > > > "Let f be an arbitrary function. There does not exist a bijective > > mapping g:N->M, where M is the set consisting of all the finite subsets > > of N together with the set K={k e N: k /e f(k)}." > > > > This is false. There always will exist such a bijection g. It will of > > course be different to f. > > Except that if the function is require to have K as a value, the f and K > and M cannot exist: > if f(n) = K = {k e N: k /e f(k)} is n a member of K or not? This is a perfectly good proof that K cannot be in the range of f. But as far as I can tell, there would still be a bijection g:N->M, and K would be in the range of g. g would of course be different from f. No? > > > > > > > > > > > > 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. > > > > > > > > > > > > > So you mean M is the set of all finite subsets of |N, together with all > > > > sets of the form K={k e |N: k /e f(k)}, where f ranges over all > > > > possible mappings |N->P(|N)? > > > > > > No. K is that single set which belongs to the function f. > > > > > > Regards, WM
From: Rupert on 21 Jun 2006 04:05 mueckenh(a)rz.fh-augsburg.de wrote: > Rupert schrieb: > > > Are you familiar with existential quantifiers and universal > > quantifiers? You are making a statement of the form "There does not > > exist an f such that...", or, alternatively "For all f it is not the > > case that..." Then later on you use f to refer to a specific function. > > You should use different letters on these two occasions. > > No. "For all f it is not the case" means the same as "there does not > exist any f for which it is the case". A f (f =/= surj) <==> ~E f (f = > sur). > I know. > And I let f be any function you like. The question is, is the second occurrence of f within the scope of the quantifier? I have given you two interpretations of what you said, one on the assumption that the second occurrence of f is not within the scope of the quantifier, and the other on the assumption that it is. And I have responded to what you said on both interpretations. > > > > "Let f be an arbitrary function. There does not exist a bijective > > mapping g:N->M, where M is the set consisting of all the finite subsets > > of N together with the set K={k e N: k /e f(k)}." > > That is what I said. > > > > This is false. There always will exist such a bijection g. It will of > > course be different to f. > > The same is true is you take Hessenberg's mapping of f:N --> P(N). It > is not surjective because of K(f). But if you define g(n+1) = f(n) and > g(1) = K(f), the proof is no longer a proof. > Yes, it is. For each mapping f:N->P(N), you have a bijection g:N->M. If you change the mapping, you've got to change the bijection. But there still is one. The proof still works. > Regards, WM
From: mueckenh on 21 Jun 2006 09:41
Rupert schrieb: > > Except that if the function is require to have K as a value, the f and K > > and M cannot exist: > > if f(n) = K = {k e N: k /e f(k)} is n a member of K or not? > > This is a perfectly good proof that K cannot be in the range of f. But > as far as I can tell, there would still be a bijection g:N->M, and K > would be in the range of g. g would of course be different from f. No? 2) 0.12324389 3) 0.23123123 4) 0.85348714 ... 1) 0.244... This is a perfectly good proof, that a surjective mapping g: |N --> [0,1] does exist, isn't it? Regards, WM |