An uncountable countable set
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From: Virgil on 22 Jun 2006 19:28 In article <1150975362.277573.39860(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > So the resolution is that either f, K_f and M_f are self-contradictory > > and do not exist at all > > if f is defined to be a surjection. That is right. And why do you > believe that a self contradictory approach should prove anything (for > instance about the surjectivity of certain mappings)? I believe that what does not exist cannot be used as an example of something that does exist. If one requires that f(n) = K(f) = {x : x not in f(x)} then f:N --> M(f) does not even exist nor does K(f) or M(f). if one does not require that f(n) = K(f) = {x : x not in f(x)} then f:N --> M(f) exists, but is not a bijection. However in that case, N and M are easily bijected, so that M is countable
From: Virgil on 22 Jun 2006 19:30 In article <1150975587.924677.62910(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1150957437.491685.293410(a)g10g2000cwb.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > > > > > > > A *surjective* mapping does not exist. The same does Hessenberg. > > > > > > > > A *surjective* mapping does exist, but it is not f. There exist a > > > > surjective mapping g -> M(f). > > > > > > Like this one?: > > > > > > 2) 0.12324389 > > > 3) 0.23123123 > > > 4) 0.85348714 > > > 5) 0.11133333 > > > 6) 0.31415161 > > > .. > > > > > > 1) 0.24446... > > > > > > This is the proof, that the proof that a surjective mapping f: |N --> > > > [0,1] does not exist, does not exist, isn't it? > > > > This is plain nonsense. > > Maybe. It is, however, the same approach as yours. If it is impossible > to find a direct surjective mapping, then first define the set and then > try to find a surjection. I can proudly declare in your words: > > A *surjective* mapping does exist, but it is not f. There exists a > surjective mapping g -> M(f). If the function f and the set M(f) exist it is because there is no n in N for which f(n) = K(f), and in that case, there are bijections between N and M(f), but f is not one of them.
From: Virgil on 22 Jun 2006 19:36 In article <1150976208.140566.164550(a)r2g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Here is another one: Let {q_1, q_2, q_3, ...} the well-ordered set of > all rational numbers. > If you say that it exists, then I can prove that it can be ordered by > magnitude without destroying the well-order. Obviously that is > impossible. Hence {q_1, q_2, q_3, ...} does not exist. I doubt that Muecken prove much of anything even when the thing is provable, but I challenge him to provide a valid proof, based only on what is deriveable from ZFC or NBG, of his claim that given any well ordering of the rationals, Muecken can prove that the standard ordering is also a well ordering. If HMuecken could do this, it would make as big a splash as the proof of FLT.
From: Virgil on 22 Jun 2006 19:38 In article <1150976829.754134.28000(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > No more than it proves Muecken does not exist. > > Please note: My name is Mueckenheim or short WM. It is fairly long, > but so much time must be for polite people. And I am not going to talk > to impolite people. > > Regards, WM If your address is mueckenh(a)rz.fh-augsburg.de, I reserve the reight to address you as mueckenh.
From: Virgil on 22 Jun 2006 21:13
In article <1151009524.179157.142240(a)r2g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Jens Kruse Andersen schrieb: > > > mueckenh wrote: > > > Jens Kruse Andersen schrieb: > > > > > > > By definition, an infinite set M is uncountable if and only if: > > > > There is no bijective mapping f from N to M. > > > > > > > > Note that M is given first, and *then* it is said that no bijective f > > > > exists for that M, with no other condition on f than it has to be > > > > bijective. > > > > > > Oh, I see. Let's apply this new insight: > > > > > > 2) 0.12324389 > > > 3) 0.23123123 > > > 4) 0.85348714 > > > 5) 0.11133333 > > > 6) 0.31415161 > > > .. > > > > > > 1) 0.24446... > > > > > > So this is the proof, that the proof that a surjective mapping f: |N > > > --> [0,1] does not exist, does not exist, isn't it? > > > > As others have said: Nonsense. A small list of numbers with no definitions, > > assumptioms, explanations or anything else is, well, a small list of > > numbers. > > I think everyone her does know this list of numbers. It is not a small > list, but a countably infinte one. (One point was missing, it should > read "...".) > > > Your final claim is hard to parse but if you think your small list of > > numbers > > with no comments proves that Cantor's diagonal proof is wrong, then: > > Nonsense. > > Comment: The first mapping read f = 1), 2), 3), .... The second mapping > reads 2), 3), 4), ..., 1). Isn't that enough? > > > > After seeing your replies here and elsewhere, I have decided to not spend > > more > > time on this discussion. > > Others have already said several times what I would say but you appear to > > ignore it. I suggest rereading the replies carefully. > > I have never stated the following problem. So you might be interested: > > Cantor said: A well-ordered set remains well-ordered, if finitely many > or infinitely many transpositions are executed. Source??? > Let's see what happens. > > Let {q_1, q_2, q_3, ...} be the well-ordered set of all positive > rationals. > 1) Consider for n e |N the rationals q_n and q_n+1. If q_n > q_n+1, > exchange q_n and q_n+1, otherwise let the succession unaltered. This > requires at most aleph_0 transpositions. > 2) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2, > exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This > requires at most aleph_0 transpositions. > 3) Consider for n e |N the rationals q_n and q_n+1. If q_n > q_n+1, > exchange q_n and q_n+1, otherwise let the succession unaltered. This > requires at most aleph_0 transpositions. > 4) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2, > exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This > requires at most aleph_0 transpositions. If each of the above steps is alleged to preserve the well-ordering, no sequence of such operations can make an ordering in which a non-empty set fails to have a smallest member according to its current ordering. But for every real number x, the set of rationals greater than x in the standard rational ordering is a non-empty set with no smallest member. Thus there are uncountably many proofs that the set of positive rationals in its standard order is not well-ordered. > > Of course, this all is defined in zero time, like the definition of > Cantor's diagonal. > After aleph_0 * aleph_0 = aleph_0 transpositions, the set {q_k_1, > q_k_2, q_k_3, ...} of all positive rationals is well-ordered and > simultaneously ordered by magnitude Claimed but not proven. |