An uncountable countable set
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From: mueckenh on 26 Jun 2006 10:07 Daryl McCullough schrieb: > >> How are you saying anything different from that? But the > >> latter statement immediately implies the statement > >> "forall f from N to P(N), f is not surjective". > > > >forall f from N to any set, K(f) is not in the image of f . > > Yes, that's true. And that implies that "forall f from N > to P(N), f is not a surjection". It is not a problem of surjectivity. This set K may be in P(|N), but it very *defintion* is nonsense. Define a bijective mapping from {1, a} on P({1}) = {{}, {1}}. a is not a natural number. There are two bijections possible. The set K cannot be mapped by a number although K is in the image of both the possible mappings. f: 1 --> {1} and a --> { } with K_f = { }, g: 1 --> { } and a --> {1} with K_g = {1}. Here we have certainly no problem with lacking elements. Nevertheless Hessenberg's condition cannot be satisfied. The set {K_f, m, f} is an impossible set. Regards, WM
From: mueckenh on 26 Jun 2006 10:20 Dik T. Winter schrieb: > In article <1151079833.849194.73370(a)c74g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > A *surjective* mapping does exist, but it is not f. There exists a > > > > surjective mapping g -> M(f). > > > > > > Indeed. But your short list does not show anything at all, I wonder what > > > I have to do with it. I think you intend to show something about the > > > diagonal number of a list of reals. But we were talking about a mapping > > > from N to P(N). So please keep to the subject. > > > > Don't you recognize that K(f) takes exactly the same function as the > > diagonal number D in Cantor's second argument? f enumerates the list > > numbers. If f can arbitrarily be replaced by g, then this proof is > > invalid and shows only what it does show in fact, namely the > > countability of all list numbers including all diagonal numbers which > > can be constructed. > > I have diffculty making sense from this. But let me try, and the answer > is (in my opinion) no. > > In the case of K(f) and M(f) I start with a bijection between N and S, where > S is the set of finite subsets of N. You state: but K(f) is not in it. I > answer: no obviously not, K(f) is not in S, and I constructed f as a > bijection between N and S. That is the same as enumerating by f the numbers of Cantor's list, which could , e.g., contain all terminating rationals. > Not you state: so M(f) is not countable, and > I say, but it is, and I construct a bijection (g) between N and M(f). > So you utter, but K(g) is not in it, and I answer, no obviously, g was a > bijection between N and M(f) and K(g) is not an element of M(f). Next > you state, but g is not a bijection between N and M(g), and my answer is: > yes, that is obvious, because g is not even a mapping between N and M(g) > because K(f) (which is in the image of g) is not an element of M(g). > Let's be generous and let you have said that it was not a bijection between > N and M(f, g) (where M(f, g) is appropriately defined). And I state, of > course not, that was not the requirement, the requirement was a bijection > between N and M(f). So each time I come with a bijection you give a > different set to which it is not a bijection. That is obvious. And each time I implement the diagonal number K(f) into the list f (insert it before the first ordinary list number, for instance), you will construct another diagonal number K(g) and say, look here, this diagonal is not contained in your g. > > Now consider the case of a bijection between N and P(N). You state: f is > a bijection, and I state: no, it is not a bijection because H(f) (H > standing for Hessenberg set) is not in the image, while it is in P(N). It may be *as a set* in P(N), but it is not in P(N) by that very definition of Hessenberg's. > OK you state, lets reformulate to g such that H(f) is in the mapping, and > I say, no, now H(g) is not in the image. You can continue, but everytime > I can state a set that is not in the image and should be there. Note that > in this case the target does *not* change. Consider one more time the bijective mapping from {1, a} on P({1}) = {{}, {1}}. a is a symbol but not a number. There are two bijections possible. The set M of all numbers which are non-generators cannot be mapped by a number although M is in the image of both the possible mappings. f: 1 --> {1} and a --> { } with M_f = { }, g: 1 --> { } and a --> {1} with M_g = {1}. Here we have certainly no problem with lacking elements. Nevertheless Hessenberg's condition cannot be satisfied. The set {M_f, m, f} is an impossible set. Regards, WM
From: mueckenh on 26 Jun 2006 10:27 Dik T. Winter schrieb: > In article <vmhjr2-F4EEE8.16301223062006(a)news.usenetmonster.com> Virgil <vmhjr2(a)comcast.net> writes: > > In article <e7hk0j01rlj(a)drn.newsguy.com>, > > stevendaryl3016(a)yahoo.com (Daryl McCullough) wrote: > > > mueckenh(a)rz.fh-augsburg.de says... > > > >Virgil schrieb: > > > > > > > >> > Cantor said: A well-ordered set remains well-ordered, if finitely many > > > >> > or infinitely many transpositions are executed. > ... > > > >If actual infinity does exist, then also an actually infinite set of > > > >transpositions must exist. Cantor knew that. > > > > > > The issue is whether "A well-ordered set remains well-ordered, if ... > > > infinitely many transpositions are executed." That's a provably false > > > statement, and I doubt that Cantor ever said it. > > > > Actually, even if it IS true, it would not allow transmutation of a set > > with a first element into one without a first element, at least if the > > transpositions are applied sequentially, as they must be. > > Mueckenheim's statement was a complete misreading of the source he quoted > in another article. I abviously can make jokes about my ability to read > German and his ability to do so, but the quote shows (without a doubt) This is obviously wrong. May it be that your lacking doubts in set theory are as well founded? > that Cantor wrote that an arbitrary number of interchanges of a > well-ordered set would not change *the number of elements* in such a set. > It does *not* state that well-orderedness is preserved. > > What I think this means (and I am not very deep in set-theory) is that > it is possible to compare the number of elements in well-orderable sets, > and so also possible to assign cardinal numbers to such sets. I have pointed out in a previous posting hat your interpretation is wrong. Cantor's "Anzahl" is connected with well-ordering and has nothing to do with cardinality. "Die Anzahl einer wohlgeordneten Menge ist also ein Begriff der in Beziehung steht zu ihrer Anordnung; bei endlichen Mengen findet er sich offenbar als unabhängig von der Anordnung; dagegen jede unendliche Menge verschiedene Anzahlen im Allgemeinen hat, wenn man sie auf verschiedene Weise als "wohlgeordnete" Menge denkt." But whether or not this is accepted: If infinity exists, then also infinitely many transpositions do exist, and none of them will destroy the well-ordering. Regards, WM
From: mueckenh on 26 Jun 2006 10:53 Daryl McCullough schrieb: > mueckenh(a)rz.fh-augsburg.de says... > > >Daryl McCullough schrieb: > > >> I'm not asking about Cantor's proof. I'm asking you to demonstrate > >> a function f from N to R such that every "individualized" real number > >> is in the image of f. Can you demonstrate such a function? If not, > >> why not? > > > >The diagonal number is not an individualized number, unless it is of > >such a simple kind I demonstrated. A random Cantor list can never be > >read to the bottom, the diagonal number can never be individualized. > > Forget about Cantor for now. I'm asking you to demonstrate a function > f from N to R such that every "individualized" real number is in the > image of f. Can you demonstrate such a function? I cannot demonstrate a function f enumerating all reals, but I can demonstrate a form containing all real numbers, the binary tree: It is briefly described in my paper http://www.fh-augsburg.de/~mueckenh/Infinity/P2%20R4%20final.pdf Regards, WM
From: Daryl McCullough on 26 Jun 2006 11:13
mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: >> Forget about Cantor for now. I'm asking you to demonstrate a function >> f from N to R such that every "individualized" real number is in the >> image of f. Can you demonstrate such a function? > >I cannot demonstrate a function f enumerating all reals So the answer is no? There is no function f enumerating all the real numbers? So the reals are uncountable. -- Daryl McCullough Ithaca, NY |