An uncountable countable set
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From: mueckenh on 24 Jun 2006 08:55 Russell schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Daryl McCullough schrieb: > > [snip] > > > > 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! > > There is *no* mapping? No mapping at all? How can that > be, if P(N) is a set with at least one member? Please tell > us more about this interesting claim of yours. There is no mapping including K as the image of at least one source element, even if surjectivity is not at all in question! Regards, WM
From: mueckenh on 24 Jun 2006 08:59 Daryl McCullough schrieb: > Yes. The reasoning is very similar: > > If f is any function from N to P(N), then > K(f) is an element of P(N) that is not in > the image of f. Therefore, f is not a surjection. > > If L is any enumeration of reals (that is, L is a > function from N to R), then D(L) is an element of > R that is not in the enumeration L. Therefore, L > is not complete. On the other hand we know that the set of individualizable real numbers is countable. > The theorem is this: > > forall A:set, > forall f: A -> P(A), > exists K: P(A), > K is not in the image of f > > which implies > > forall A:set, > forall f: A -> P(A), > f is not a surjection from A to P(A) But for all sets S constructed by the diagonal argument, we find a surjective mapping g: |N --> S. > > which implies > > forall A:set, > there does not exist a surjection from A to P(A) > > >and shows only what it does show in fact, namely the > >countability of all list numbers including all diagonal > >numbers which can be constructed. > > It shows that every list of reals, there is at least one > real that is not on the list. From that, it follows that > there is no list containing all real numbers. No. It means that there is a paradox which can be remedied by switching the mapping from f to g and this is possible for every real number constructed. Therefore there is no uncountable set of reals. > By definition, > that means that the reals are uncountable. > > >> 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! > > Well, there certainly is a mapping from N to P(N). For > example, let f(n) = { n }. So I don't know what you could > possibly mean by saying that there is no mapping. There is no mapping including K as the image of at least one source element, even if surjectivity is not at all in question! Regards, WM
From: mueckenh on 24 Jun 2006 09:00 Daryl McCullough schrieb: > >The assumption includes that |N and P(|N) do exist as complete sets. > > Actually, the nonexistence of a mapping from N to P(N) has nothing > to do with whether N and P(N) exist as "complete sets". If you > want to think of them as potentially infinite, that's fine. It's > still possible to have a bijection between two potentially infinite > sets. Correct. If |N and P(|N) are potentially infinite, then the definition of a mapping doesn't cause any problem, but it can be shown bijective, because there is no completeness. > > The function f(x) = 2*x is a surjection from the set of naturals > to the set of even naturals. It doesn't matter whether you think > of the naturals as "completed" or not; if x is any natural whatsoever, > I can multiply by 2 to get an even natural. And given any even natural, > I can divide by two to get a natural. It doesn't matter whether the > naturals are "complete" or not. > > Similarly, there is a surjection from P(N) to [0,1]. Given any set S of > naturals, I can generate a real number as follows: > > r(S) = sum over all j in S of 2^{-j} > > I don't need to think of the set [0,1] as "completed" in order to > understand that this mapping is surjective. "Surjective" means that all elements are in the image. A potentially infinite set is always finite though surpassing any given limit. Therefore the question of all of them cannot be decided. Regards, WM
From: mueckenh on 24 Jun 2006 09:02 Virgil schrieb: > 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. The transpositions are just as sequentially as the construction of the diagonal number. > There would have to be a first transposition in the sequence which > effects the change, and a single transposition is incapable of doing > this. This arguing stems from König (1905). We know that there are only countably many constructible reals. If there is a well-ordering of the reals, then within this order there must be a first non-constructible real. This is a paradox. Hence, there is no well order. Regards, WM
From: Daryl McCullough on 24 Jun 2006 09:30
mueckenh(a)rz.fh-augsburg.de says... >"Surjective" means that all elements are in the image. A potentially >infinite set is always finite though surpassing any given limit. >Therefore the question of all of them cannot be decided. That's not true. If A is a "potentially infinite" set, then saying "forall x in A, Phi(x)" just means that if x is any thing of type A, then Phi holds of x. It doesn't matter whether A is "completed" or not. We know perfectly well that if x is a natural number, then x+x is an even natural number. From this fact, it follows that forall x: Natural, x+x is even "forall" doesn't mean that anyone has *checked* every element of the set. -- Daryl McCullough Ithaca, NY |