An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 23 Jun 2006 14:23 In article <1151079612.527394.67300(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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 erl?uternden Anmerkungen sowie > mitmErg?nzungen 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 ge?ndert wird, durch welche nicht, l??t sich einfach so > beantworten, da? diejenigen und nur diejenigen Umformungen die Anzahl > unge?ndert lassen, welche sich zur?ckf?hren 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. Not at all. In order to be unambiguous transpositions must be carried out sequentially, meaning that only countably many can be applied. Mueckenh's analysis is not unambiguously a sequence of transpositions. Also he does not show that any single transposition can re-order a set having no first member into one which does have a first member. Similarly in the opposite direction, no single transposition will cause a set with a first element to become one without a first element. Thus his alleged proof fails.
From: Virgil on 23 Jun 2006 14:33 In article <1151079756.995887.177600(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. One can use either theorem to prove the other's conclusion, but Cantor's first proof relies on the topological properties of the reals and connected subsets of the reals, so is essentially topological, whereas the proof that there is no surjection from N to P(N) is purely algebraic. > > > [...] > > > > > > 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. The only disproof of the uncountability of, say P(N), would be the actual construction of a surjection from N to P(N), or at least proof that such a construction was possible. Your move, Mueckenh.
From: Virgil on 23 Jun 2006 14:42 In article <1151079833.849194.73370(a)c74g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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 By definition, a set M is countable if and only if there SOME surjection g:N --> M exists and is uncountable if no such surjection can exist. This means that, given a set M, one must allow ANY function from N to M to be considered. So that given that M(f) exists at all, one is not constrained to only the function f which defines K(f) and M(f), but can consider whatever functions imaginable from N to M(f). Since it is easy to construct surjections from N to the set of all finite subset os P(N), it is also easy to do it from N to M(F).
From: Virgil on 23 Jun 2006 14:52 In article <1151080005.054210.206040(a)r2g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Let M(f) be any set containing K(f) as an element, and consider all possible functions f: N --> M(f). Is it possible for there to be any n in N such that f(n) = K(f) = {x in N:x not in f(x)} ? A moments thought shows it to be impossible. Thus the very existence of f, K(f) and M(f) is conditional upon K(f) NOT being a value of f, thus upon f NOT being surjective. The number of elements in M(f), as long as it is more than just K(f), is immaterial. If f is to exist at all, it is not a surjection.
From: Virgil on 23 Jun 2006 15:02
In article <1151080124.494952.93940(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > > >This is proven by three invalid proofs. > > > > We're only discussing one right now: > > 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. > > > If f is any function from > > N to P(N), then let K(f) = { x in N | x is not in f(x) }. Then > > > > 1. K(f) is a subset of N. > > 2. K(f) is an element of P(N). > > 3. K(f) is not in the image of f. > > 4. Therefore, f is not a surjection. > > > > That's all there is to it: If f is any function from N to P(N), > > then f is not a surjection from N to P(N). That is logically > > equivalent to "There is no surjection from N to P(N)", which > > is by definition what "P(N) is uncountable" means. > > > > >It is wrong, because all it shows is the construction > > >of a number of a set of countably many constructible numbers. > > > > What it shows is that the assumption that f is a surjection > > from N to P(N) leads to a contradiction. If something leads > > to a contradiction, then it is provably false. > > So drop this assumption. There is no mapping, even if surjectivity is > not at all in question! > > Regards, WM Mueckenh is WRONG! If surjectivity is not required, all sorts of f's. can exist! Consider the following: f: N --> P(N) : f(n) = {} for all n in N. Then the image of f is {{}} Then K(f) = {x in N:x not in f(x)} = N, which shows f is not a surjection, but can exist if not required to be surjective. So there IS a mapping in this case, and Mueckenh is wrong again. |