Why can no one in sci.math understand my simple point?
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From: WM on 17 Jun 2010 04:07 On 17 Jun., 05:26, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-15, Peter Webb <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > No. You cannot form a list of all computable Reals. If you could do > > this, then you could use a diagonal argument to construct a > > computable Real not in the list. > > You can form a list of all computable reals (in the sense of > mathematical existence). However, such a list is not itself > computable. That does not matter. There exists a list containing all computable reals in all possible languages. Therefore the set of reals that can serve as doiagonals of a Cantor list is countable. Regards, WM
From: Tim Little on 17 Jun 2010 04:49 On 2010-06-17, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Virgil" <Virgil(a)home.esc> wrote in message >> An infinite set is defined to bee countable if and only if there is a >> surjection from the set of natural numbers to that set. When such a >> function is a bijection, it is commonly called a list. > > Only if the bijection can be explicitly created. You apparently have some bizarre private definition of "list". Explicitness has nothing to do with it. Though even if it did, you are incorrect. Given any numbering of Turing machines, a list of computable reals ordered by the least-numbered Turing machines that compute them is quite explicit. The practical difficulties of establishing which Turing machines halt, which are equivalent and so on are just that: practical difficulties which have nothing to do with mathematical theory. - Tim
From: WM on 17 Jun 2010 05:06 On 17 Jun., 09:54, Virgil <Vir...(a)home.esc> wrote: > In article <4c19cd2c$0$316$afc38...(a)news.optusnet.com.au>, > "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > Cantor's proof applied to computable numbers proves you cannot form a > > computable list of computable numbers. Cantor's proof applied to Reals > > proves you cannot form a computable list of Reals. Cantors proof is nonsense from the beginning, because a real number can never be defined by an infinite sequence alone. A definition defines something, but an infinite sequence does not define a number before the last digit is known. > > To be correct, there is no computable list of ALL of the computable > numbers, even though the set of computable numbers is e countable, but > there are lots of possible computable lists of computable numbers. And there are lots of lists of more than all computable numbers, namely lists of all finite expressions. Regards, WM
From: Aatu Koskensilta on 17 Jun 2010 06:16 Virgil <Virgil(a)home.esc> writes: > In article > <995d761a-f70f-4bca-b961-8db8e1663e3f(a)d37g2000yqm.googlegroups.com>, > WM <mueckenh(a)rz.fh-augsburg.de> wrote: > >> On 15 Jun., 22:24, Virgil <Vir...(a)home.esc> wrote: >> >> >> > Note that it is possible to have an uncomputable number whose >> > decimal expansion has infinitely many known places, so long as it >> > has at least one unknown place. >> >> That is mathematically wrong. > > It may not match every definition of 'uncomputable', but otherwise it is > right. It doesn't really match any definition of 'computable'. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Sylvia Else on 17 Jun 2010 09:56
On 15/06/2010 2:13 PM, |-|ercules wrote: > Consider the list of increasing lengths of finite prefixes of pi > > 3 > 31 > 314 > 3141 > .... > > Everyone agrees that: > this list contains every digit of pi (1) > > as pi is an infinite digit sequence, this means > > this list contains every digit of an infinite digit sequence (2) > > similarly, as computable digit sequences contain increasing lengths of > ALL possible finite prefixes > > the list of computable reals contain every digit of ALL possible > infinite sequences (3) Obviously not - the diagonal argument shows that it doesn't. > > OK does everyone get (1) (2) and (3). > No. Sylvia. |