Why can no one in sci.math understand my simple point?
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From: Daryl McCullough on 21 Jun 2010 06:41 Peter Webb says... >Cantor constructs the antidiagonal by taking the list and then computing a >missing Real based upon the decimal expansion of every Real on the list. That's right, so Cantor showed how to compute an antidiagonal using the list of reals as an input. That shows that the antidiagonal is computable *relative* to the list. It does not show that the antidiagonal is computable. "Computable relative to a list" and "computable" are two different things. This is a pretty simple concept, Peter. Mathematically, you have a function antidiagonal(L,n) which returns the nth digit of the antidiagonal, given the list L and n. That function is computable. But for the antidiagonal to be computable (not computable *relative* to L), you would have to be able to come up with a function antidiagonal(n) that, for *any* n, gives the nth digit of the antidiagonal. Two different concepts: computable relative to L, and computable. -- Daryl McCullough Ithaca, NY
From: WM on 21 Jun 2010 07:06 On 21 Jun., 12:41, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Peter Webb says... > > >Cantor constructs the antidiagonal by taking the list and then computing a > >missing Real based upon the decimal expansion of every Real on the list. > > That's right, so Cantor showed how to compute an antidiagonal > using the list of reals as an input. That shows that the antidiagonal > is computable *relative* to the list. It does not show that the > antidiagonal is computable. "Computable relative to a list" and > "computable" are two different things. > > This is a pretty simple concept, Peter. Mathematically, > you have a function antidiagonal(L,n) which returns the > nth digit of the antidiagonal, given the list L and n. > That function is computable. But for the antidiagonal > to be computable (not computable *relative* to L), you > would have to be able to come up with a function > > antidiagonal(n) > > that, for *any* n, gives the nth digit of the antidiagonal. > > Two different concepts: computable relative to L, and computable. But none of them does help to save set theory. The diagonal numbers computable relative to a list are countable. And so are the diagonal numbers computable relative to a list that is extended in the n-th step by the diagonal number computed in step n - 1. Nevertheless, there is not list of these "relative to the list" computable diagonal numbers. Therefore Cantor's proof shows the uncountability of a countable set. Regards, WM
From: Jesse F. Hughes on 21 Jun 2010 08:46 Newberry <newberryxy(a)gmail.com> writes: > What I had in mind is that if you drop the axiom of extent you can > either conclude that there is no such list or that the anti-diagonal > does not exist analogously to the conclusion that the set R = {x | ~(x > in x} does not exist. I'm not sure what the axiom of extent is. > BTW, Cantor did not work in ZFC and the argument given by McCullough > appeared to be on the intuitive level rather than in any particular > formal theory. I rather suspect that Daryl had ZFC in mind, but again, we can ask him to clarify. > > I will grant you that Zermelo figured how to construct a system where > the diagonal argument goes through and his system is probably > consistent. It is a nonsense nevertheless. All the transfinite set > theory is a completely superfluous metaphysics. As long as you agree that the theorem is valid in ZFC, I don't see that we disagree. So, in your opinion, ZFC is a bad theory. Yeah, okay, I don't see a point in arguing that claim. -- "Looking at their behavior I see them endangering not only their own futures, but that of their families, and now, considering my latest result, the future of people all over the world." -- James S. Harris, on the shortsightedness of his mathematical critics
From: Daryl McCullough on 21 Jun 2010 09:03 Jesse F. Hughes says... >Newberry <newberryxy(a)gmail.com> writes: >> BTW, Cantor did not work in ZFC and the argument given by McCullough >> appeared to be on the intuitive level rather than in any particular >> formal theory. > >I rather suspect that Daryl had ZFC in mind, but again, we can ask him >to clarify. What I prefer is to work in typed first-order logic, where there are types for naturals, ordered pairs, functions, sets, etc, and where quantifiers are all relativized to a type. Such an enriched language can be translated into the language of ZF, although in my opinion, the translated forms of sentences are harder to understand than the type-theoretic versions. -- Daryl McCullough Ithaca, NY
From: Sylvia Else on 21 Jun 2010 09:05
On 21/06/2010 7:51 PM, WM wrote: > On 20 Jun., 22:18, Virgil<Vir...(a)home.esc> wrote: >> In article >> <f92c169d-ee85-40c2-aa82-c8bdf06f7...(a)j4g2000yqh.googlegroups.com>, >> >> WM<mueck...(a)rz.fh-augsburg.de> wrote: >>> On 20 Jun., 17:51, "Jesse F. Hughes"<je...(a)phiwumbda.org> wrote: >> >>>> Cantor was this utterly insane freak who chose not to accept Newberry's >>>> word for it, and instead *prove* that there was no list of all real >>>> numbers. Obviously, his proof is nonsense, because, after all, Newberry >>>> said there was no list. >> >>> His proof is nonsense because it proves that a countable set, namely >>> the set of all reals of a Cantor-list and all diagonal numbers to be >>> constructed from it by a given rule an to be added to this list, >>> cannot be listed, hence, that this indisputably countable set is >>> uncountable. >> >> That is a deliberate misrepresentation of the so called "diagonal proof". > > But this proof can be applied to this countable set and shows its > uncountability. > Let's see. How might that work... OK, by way of example, take the set of rationals. It's countable, so we can list them. Now construct an anti-diagonal. It's clearly not in the list, so... .... um, well what, exactly? It should be in the list if, and only if, it's rational. But unless you can prove that it is rational, you have nothing. It seems more than likely that the anti-diagonal is irrational and isn't in the list for the good reason that it doesn't belong there. Sylvia. |