Why can no one in sci.math understand my simple point?
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From: WM on 15 Jun 2010 11:27 On 15 Jun., 13:53, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > WM <mueck...(a)rz.fh-augsburg.de> writes: > > In what form does an uncountable real exist? > > What is an "uncountable real"? Sorry. Uncomputable real was meant. (But perhaps it's the same, because to be countable implies to exist?) Regards, WM
From: WM on 15 Jun 2010 11:31 On 15 Jun., 14:45, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > news:62ae795b-1d43-4e1f-8633-e5e2475851aa(a)x21g2000yqa.googlegroups.com... > > > On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >> (B) There exists a real number r, > >> Forall computable reals r', > >> there exists a natural number n > >> such that r' and r disagree at the nth decimal place. > > > In what form does r exist, unless it is computable too? > > Of course its computable. > > You compute it by changing the nth digit of the nth number on the list to a > 7, unless it is already a 7 in which case you make it an 8. > > Sounds pretty easy to compute, I would have thought. I reckon I could code > it in about 2 minutes. > > > But if r is computable, then this theorem shows that the computable > > numbers are uncountable. Contradiction. > > Huh? > > > And if r is not computable, then it is impossible to prove > > disagreement with any r'. > > Computing the diagonal number is actually very easy. Therefore the diagonal number is a computable real. Then Cantor's proof works exclusively on a countable set, namely the set of computable reals. And it shows that this set is uncountable. This result is wrong. Regards, WM
From: WM on 15 Jun 2010 11:38 On 15 Jun., 16:06, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > WM says... > > > > >On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >> (B) There exists a real number r, > >> Forall computable reals r', > >> there exists a natural number n > >> such that r' and r disagree at the nth decimal place. > > >In what form does r exist, unless it is computable too? > > r is computable *relative* to the list L of all computable reals. > That is, there is an algorithm which, given an enumeration of computable > reals, returns a real that is not on that list. > > In the theory of Turing machines, one can formalize the notion > of computability relative to an "oracle", where the oracle is an > infinite tape representing a possibly noncomputable function of > the naturals. We should not use oracles in mathematics. A real is computable or not. My list contains all computable numbers: 0 1 00 .... This list can be enumerated and then contains all computable reals. But I understand that, in order to escape contradicitons, you must distinguish between "computable" and "computable relative" of first class and "coputable relative" of second class and so on, because the idea of uncomputable reals cannot be used to avoid the contradiction of Cantor's theory. Regards, WM
From: WM on 15 Jun 2010 11:42 On 15 Jun., 16:17, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > In this sense, the antidiagonal of the list of all computable reals > is definable (but not computable). That is nonsense. To define means to let someone know the defined. If he knows it, then he can compute it. The set of definitions and therefore the set of definable reals is countable too. So this relativation does not help save set theory. Regards, WM
From: WM on 15 Jun 2010 11:43
On 15 Jun., 16:18, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Peter Webb says... > > >"WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > >news:62ae795b-1d43-4e1f-8633-e5e2475851aa(a)x21g2000yqa.googlegroups.com... > >> On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >>> (B) There exists a real number r, > >>> Forall computable reals r', > >>> there exists a natural number n > >>> such that r' and r disagree at the nth decimal place. > > >> In what form does r exist, unless it is computable too? > > >Of course its computable. > > No, it's computable *relative* to the list of all computable reals. > But that list is not computable. That is nonsense! The list of all definitions is possible and obviously contains all definitions of real numbers. Regards, WM |