An uncountable countable set
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From: Russell on 23 Jun 2006 15:15 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.
From: Daryl McCullough on 23 Jun 2006 16:28 mueckenh(a)rz.fh-augsburg.de says... >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? 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. >f enumerates the list numbers. If f can arbitrarily be >replaced by g, then this proof is invalid On the contrary, the proof has the form of a universal statement: It is true for all f that f is not surjection from N to P(N). If it's true for all f, then it is true for g as well. 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) 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. 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. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 23 Jun 2006 16:39 mueckenh(a)rz.fh-augsburg.de says... >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? "Surjective" is a 3-place predicate: f is a surjective map from A to B. It doesn't make sense to say that f is surjective without specifying the range B. What set A are you talking about? What set B are you talking about? If (1) A and B are any sets whatsoever, (2) f is any function from A to B, (3) K(f) is defined to be { x in A | x is not an element of f(x) }, and (4) K(f) is an element of B, then it follows that (5) f is not a surjection from A to B -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 23 Jun 2006 16:47 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. >> >> Source??? > >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. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 23 Jun 2006 17:01
mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: > >> If an assumption leads to a contradiction, then that assumption must >> be false. > >The assumption includes that |N and P(|N) do exist as complete sets. Perhaps, but that assumption doesn't lead to a contradiction. -- Daryl McCullough Ithaca, NY |