An uncountable countable set
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From: Dik T. Winter on 20 Jun 2006 20:19 In article <1150836068.607034.205340(a)y41g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > In order to define K_f you first have map all natural numbers n of |N > on the finite subsets of |N. Then consider K_f. But there is no element > k of |N remaining, which could be mapped on K. Hence the set M is > uncountable. Wrong. We can construct a bijection between N and M_f, but it will only not be f. And do not tell me that M_f changes when I construct that other mapping. Bijections are not done with moving targets. > It is about the same as Cantors diagonal proof. The > diagonal number depends on all numbers of his list. Moreover, after > constructing the diagonal number, it turns out to belong to a countable > set. But that is a different list. You do not understand the proof. Given *any* list it is possible to construct a real number that is not in that particular list. That it is in a different list is irrelevant. That is wat you always do. You move the target each time. The proof just shows that there is *no* list that is complete. Otherwise, give me a *complete* list. I can generate a number that is not in that list, so the list is not *complete*. I think you have problems with proofs by contradiction. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 20 Jun 2006 20:30 In article <1150836393.730055.231800(a)y41g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > That raises the question of whether such a set as M_f can exist at all. > > > > If it cannot exist, then the issue of whether a non-existent set is > > countable is irrelevant. > > Of course it exists. There are all finite subsets of |N. And there is, > for any function f, the set of all natural numbers, which are not > mapped on sets containing them. If |N exists, then K exists too. Maybe > |N does not exist completely? But K(f) depends on f, so M(f) depends on f. This does not preclude a bijection between N and M(f). It only tells us that f is not a bijection. Now construct a bijection between N and M(f) and call it g. It clearly is a bijection between N and M(f), so M(f) is countable. You can not add *now* to the conditions that it was not M(f) to which should be mapped but M(g). That is cheating. g is a proper bijection between N and M(f). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 20 Jun 2006 20:33 In article <1150836884.833877.138440(a)c74g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > M is countable means: There is an enumeration of the elements of M, > i.e., a bijective mapping |N <--> M. But there is *no* M. There is only a M(f). Pray remain a bit consistent. M and K depend on a mapping f. > If |N does exist, then each of its subsets must exist, including K. > What does not exist is the bijection |N <--> M. There is *no* M. You state here is no bijection between N and M(f). The only thing you have proven is that *f* is not such a bijection. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 20 Jun 2006 21:16 In article <1150835899.493068.116800(a)g10g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1150710209.401911.130270(a)i40g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > 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 ? > > > > As soon as you tell us that M_f exists at all, YOU are telling us that > > there must be such a set, K. The only other option is that there is no > > such set K and no such set M_f and no such function f at all, in which > > case the uncountable-countable-set becomes a non-existent > > uncountable-countable-set. > > Of course K exists for every mapping f (if |N exists). But the set {f, > k, K} is a paradox set. If K_f exists it is not paradoxical and if it is paradoxical, it does not exist. The paradox occurs if one has any function f with domain N and image some subset of P(N) containing K_f. If the image of f is not required to contain K_f, there re no problems. > It is not suitable to show that a bijection |N > --> {all finite subsets of P(|N) plus one further set} does not exist. > Why do you believe it could be capable of showing that |N --> P(|N) > does not exist? It is your example! Work it out for yourself why f(n) = {m in N: m not in f(m)} doesn't work.
From: Dik T. Winter on 20 Jun 2006 21:10
In article <1150837769.423770.307920(a)u72g2000cwu.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > 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. It contains all finite subsets of |N and that set > which contains all natural numers wich under f are mapped on sets not > containing them. If |N is a set, then M is clearly a set too. Yes, M(f) is, for each and every f: N -> S (the set of finite subsets of N), M is not. entier(x) is an integer for each real x, that does not mean that entier is an integer, it is a function. And so your M is a function, dependent on its argument. > > And in order to tell whether that is > > uncountable or not you have to first define cardinality on such > > non-sets. > > Cardinality is aleph_0 is there is a bijection with |N. If that is > impssible, then cardinality is larger. Yes. But you first have to show that M is a set. Not that M(f) is a set for each mapping f. > > > That is not at all the way a bijection has to be constructed between |N > > > and M. That is simply impossible! > > > > 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. > > A bijection can be constructed from |N to the numbers of Cantors list > and the diagonal number. You only first have to construct the latter. You are putting it backwards. The proof is that *given any list* it is possible to construct a number not on the list. But good luck at constructing your bijection. When are you finished? And what if a number is found not on your list (the proof shows that a number can be constructed not on the list)? > > > But if you like tricks of this kind, then you can biject |N and its > > > power set P(|N) \ K. Subsequently take > > > 1. g(0) = K(f) > > > 2. g(n) = f(n - 1) when n > 0 > > > and we see that Cantor's theorem is not proven by Hessenberg's proof, > > > because this proof makes use of the impredicable definition of a set > > > which does not and cannot exist, as well as the barber who shaves all > > > men of his village. > > > > Sorry, this is plain nonsense. You can not construct a bijection between > > P(N) \ K. > > Who told you? Remember our discussion quite some time ago? I gave an explicit proof of this. However, this is neither here nor there. We start with a source and a target. The source and target remain fixed (they are sets, you know). Your discussion about M(f) and P(N)\K all do not disprove anything, you are only moving the target. And bijections between sources and moving targets are impossible. > > The difference between this and your construction with M(f) > > is that the target M(f) depends on the bijection you are constructing. > > The same holds in case of the proof of Cantor's theorem. No, it is not. > > P(N) is a properly defined set that does not depend on anything but N. > > But the set of numbers which are mapped on sets not containing them is > not proerly defined. See, the target is P(N), it does not depend on the bijection you are constructing. It is fixed. The set of numbers which are mapped on sets not containing them is properly defined. It is an existing set and depends on the mapping. > > But you are wrong. For every mapping f, the set K does exist (and is > > an element of P(N), which does not depend on f), > > Of course it does. That is very obvious. Do you think P(N) depends on f? The set K depends on f, agreed. > > but it is not in the > > image of f, which it should be if f is a bijection. Why no remark on this? > > The triple {f, k, K} does indeed not exist, but if f is a bijection it > > should exist. It is just this contradiction that shows that f is not > > a bijection. > > Exactly the same holds in my example. Of course every subset K of |N > does exist. I is only the triple which does not. It is just this > contradiction that shows that f is not a bijection. Indeed, for f being a bijection it should exist. So f is not a bijection between N and M(f). This does not state that there is no bijection between N and M(f). There *are* bijections between N and M(f), for each and every f you can construct such a bijection, it only is not f. On the other hand, for every purported bijection between N and P(N) it can be shown that it is not a bijection. You can not do that for every purported bijection between N and M(f). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |