An uncountable countable set
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From: Virgil on 21 Jun 2006 16:00 In article <1150876556.902172.6800(a)i40g2000cwc.googlegroups.com>, 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). > > And I let f be any function you like. Let it be any function which does not have K_f as a value. > > > > "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. Then f is not arbitrary, since you insist on restricting the function f to have some n in N such that f(n) = {x in n: x not in f(x)}. With that restriction, no such f can exist, absent that restriction, lots of them do.
From: Virgil on 21 Jun 2006 16:03 In article <1150897288.785127.87000(a)g10g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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? No. And it totally ignores the fact that Muecken's alleged example of a n uncountable countable set is fatally flawed, as is shown above.
From: Daryl McCullough on 21 Jun 2006 18:24 In article <1150876556.902172.6800(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de says... > > >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: Daryl McCullough on 21 Jun 2006 18:35 mueckenh(a)rz.fh-augsburg.de says... >Rupert schrieb: >> "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. Sure it is. f is not a surjection from N -> P(N), because K(f) is not in the image of f. Similarly, g is not a surjection from N to P(N) because K(g) is not in the image of g. Let's go through this one more time. Let f be any function from N to P(N). Let image(f) = { A in P(N) | exists x in N: f(x) = A } Let K(f) = { x in N | x is not in f(x) } Let M(f) = { A in P(N) | A is in image(f) or A = K(f) } Finally, let g be defined by g(0) = K(f) g(x+1) = f(x) Okay, so we have the following facts: 1. f is not a surjection from N to M(f) 2. f is not a surjection from N to P(N) 3. g is a surjection from N to M(f). 4. g is not a surjection from N to M(g). 5. g is not a surjection from N to P(N). So, M(f) is countable, since there is a surjection g from N to M(f). But there is no surjection from N to P(N). So P(N) is *not* countable. -- Daryl McCullough Ithaca, NY
From: Rupert on 21 Jun 2006 19:33
mueckenh(a)rz.fh-augsburg.de wrote: > 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? > No, that's absolute nonsense. > Regards, WM |