An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 20 Jun 2006 16:46 Virgil schrieb: > In article <1150728692.949790.258040(a)h76g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Daryl McCullough schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de says... > > > > > > >> > 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. > > > > > > It's not exactly clear what you are talking about here. Do > > > you mean that M is the set of all finite subsets of N? > > > > Above I spelled out clearly that M contains the set of all finite > > subsets of |N and one more set K. What is unclear? > > > > > In > > > that case, there is definitely a bijection between M and N: > > > > If M was the set of all finite subsets of N, that would be easy enough > > to see. > > > > > If you let K = { k in N such that k is not an element of f(k) } > > > then we can see that > > > > > > K = N > > > > Such a mapping is possible with no doubt. There remains only one > > question: what number is mapped on K? > > 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? > So that Meucken's first duty is to prove that a function having the sort > of codomain, M_f, he describes can exist at all. > > If ever that is established, the construction of a bijection between it > and N is fairly trivial. If |N exists, then K exists too. Maybe |N does not exist completely? Regards, WM
From: mueckenh on 20 Jun 2006 16:54 Virgil schrieb: > > The element K of M does not exist. Excuse me. The set {f, k, K} does not exist. > > That requires proof, and is false in at least one case. > Let f(n) = {} for all n, then K_f is quite well defined, and equals N. You are correct. Of course K_f does exist for any f (if all natural numbers do exist). Nevertheless you see that the mapping is not a bijection beause no f(n) = K. > > > > If it is proven impossible to set up a mapping between M and |N, then M > > is uncountable. > > Wrong! M is countable means: There is an enumeration of the elements of M, i.e., a bijective mapping |N <--> M. > > If it is impossible to set up an injection from N to M_F or a surjection > from M_f to N, and the axiom of choice is assumed, only then is M_f > uncountable. > > And if K does not exist, If |N does exist, then each of its subsets must exist, including K. What does not exist is the bijection |N <--> M. Regards, WM
From: mueckenh on 20 Jun 2006 16:58 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. > > > > 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 20 Jun 2006 17:09 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. 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. > 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. > > > > 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! > > 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. > > > 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? > 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. > 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. > 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. > but it is not in the > image of f, which it should be if f is a bijection. > > > Of course, one may define a mapping from |N on P(|N), for instance n > > --> {n}, where K is well defined. But the set {f, k, K}, essential for > > Hessenberg's proof, does not exist. This shows that the customary > > proof of Cantor's theorem makes use of a set which does not exist, > > namely {f, k, K}. Therefore, its non-existence does not prove anything > > about surjectivity of mappings. > > 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. Regards, WM
From: Dik T. Winter on 20 Jun 2006 20:07
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. 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. > 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. 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |