An uncountable countable set
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From: Virgil on 20 Jun 2006 22:17 In article <1150837769.423770.307920(a)u72g2000cwu.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1150728187.749465.36710(a)g10g2000cwb.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <1150718841.726873.223390(a)g10g2000cwb.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. > > > > In that case M is not a set. > > Of course it is. But given two different functions, say f and g, one has M_f != M_g. So "M" is not a set but the value of some function whose arguments are f and g. > Cardinality is aleph_0 is there is a bijection with |N. If that is > impssible, then cardinality is larger. Or smaller, or unknown. > > > > > > 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! It is a good deal more possible that the garbage claims Meucken has been making. > > > > A bijection can be constructed, you only can not name if f. > > First you must find out what the set {f, k, K} is. *That* is > impossible, not K. F and K and M are all impossible as long as Muecken requires that f have K as a value. Absent that requirement, it is all possible and the resulting M is countable.
From: Rupert on 21 Jun 2006 02:14 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. > > > > > > 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 02:17 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? > > > > > 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. An alternative interpretation for what you're trying to say would be "It is not the case that there exists a set M of subsets of N and a bijection f:N->M, such that M consists of all the finite subsets of N together with the set K={k e N: k /e f(k)}." This is quite correct, but it does not amount to identifying an uncountable countable set. What you've done is observe that there does not exist a pair (M,f) with certain properties. To identify an uncountable set you have to first identify a set M and then observe that there does not exist an f with certain properties. > > > > > > 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: Virgil on 21 Jun 2006 02:45 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? > > > > > > > > > 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: mueckenh on 21 Jun 2006 03:39
Dik T. Winter schrieb: > > > A set is required which is the image of k if it is not the image of k. > > A barber is required who shaves himself if he does not. (In case f should be surjective.) > 1. Given a mapping f: N -> P(N), the set K(f) constructed according to > Hessenberg *does* exist. If f were a bijection it is required (by > the *definition* of bijection) that K(f) (because it is an element > of P(N)) is the image of some element of N, but it is not, showing > that f is not a bijection. P(N) does not depend on f, so there is > no mapping between N and P(N) that is a bijection. Who told you? Map f : N --> P(N) \ K(f). That has not been proven impossible. Then map g(n+1) = f(n) and g(1) = K(f). There is no proof that this cannot be a bijection. > 2. Given a mapping f: N -> S, the sets K(f) and M(f) do exist. If f > were a bijection between N and M(f), K(f) should be in the image, > it is not, so f is not a bijection. From this you can *not* > conclude that M(f) is uncountable, because there can be a bijection > g: N -> M(f). You can at most conclude that the union of *all* M(f)'s > is uncountable. And that is easy, that union is just P(N). > > > On the other hand, why should any mapping not exist? If you claim that > > a surjective mapping |N --> P(|N) must contain the set K in any case, > > then there is no arguing why my mapping should not exist. > > Your mapping does exist. See (2) above. Of course. But only if f is not surjective. And g can be applied to Hessenberg's proof in the same way. Regards, WM |