An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 23 Jun 2006 12:17 Virgil schrieb: > > 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. 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 diagonals which can be constructed. Regards, WM
From: mueckenh on 23 Jun 2006 12:20 Virgil schrieb: > > Cantor said: A well-ordered set remains well-ordered, if finitely many > > or infinitely many transpositions are executed. > > Source??? If actual infinity does exist, then also an actually infinite set of transpositions must exist. Cantor knew that. G E O R G C A N T O R: GESAMMELTE ABHANDLUNGEN MATHEMATISCHEN UND PHILOSOPHISCHEN INHALTS Mit erläuternden Anmerkungen sowie mitmErgänzungen aus dem Briefwechsel Cantor - Dedekind Herausgegeben von ERNST ZERMELO Nebst einem Lebenslauf Cantors von ADOLF FRAENKEL1966 GEORG OLMS VERLAGSBUCHHANDLUNG HILDESHEIM p. 214: "Die Frage, durch welche Umformungen einer wohlgeordneten Menge ihre Anzahl geändert wird, durch welche nicht, läßt sich einfach so beantworten, daß diejenigen und nur diejenigen Umformungen die Anzahl ungeändert lassen, welche sich zurückführen lassen auf eine endliche oder unendliche Menge von Transpositionen, d. h. von Vertauschungen je zweier Elemente." Cantor admitted infinitely many transpositions. I think he meant countably many. But more aren't needed. .. > > If each of the above steps is alleged to preserve the well-ordering, And why shouldn't it, if infinitely many steps do exist? > 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. That is obvious. Therefore we have the proof that infinity, like the set of all rationals does not exist. Regards, WM
From: mueckenh on 23 Jun 2006 12:22 Virgil schrieb: > And is that "first proof" for the theorem that there is no surjection > from N to P(N) or for the theorem that there is no surjection from N to > R, the set of all reals? The latter. It was the first proof of an uncountable set. > > They are not quite the same. But fairly the same, very fairly. > [...] > > > > It is ridiculous to believe this to be a proof of uncountability. > > It is even more ridiculous to believe otherwise. At least within ZFC or > NBG. It is ridiculous to believe in Copernicus in the frame work of the bible. Nevertheless I believe in Copernicus. Regards, WM
From: mueckenh on 23 Jun 2006 12:23 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. Regards, WM
From: mueckenh on 23 Jun 2006 12:26
Dik T. Winter schrieb: > > Because, in case f should be surjective, a number k need be contained > > in a set in which it must not be contained. > > Yes, and that shows that whatever mapping f you give, it is simply not > surjective. And if I map all natural numbers on a set with only two elements, the empty set and that single set K? The mapping is not surjective? Why should it not be *surjective*, if there are far more elements in the source than in the target? The mapping is impredicably defined. Regards, WM |