Why can no one in sci.math understand my simple point?
in [Logic]
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From: |-|ercules on 20 Jun 2010 01:57 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 20/06/2010 6:10 AM, |-|ercules wrote: > >> If all digits of a single infinite expansion can be contained with >> increasing finite prefixes, >> and the computable set of reals has EVERY finite prefix, then all digits >> of EVERY infinite >> expansion are contained. > > It's far from clear what that actually means, but in any case you > certainly haven't proved it. > > Sylvia. Hypothesis: an real number contains a finite sequence that is not computable. Contradiction Therefore: all digits of every infinite expansion are contained in the list of computable digit sequences. QED Herc
From: Sylvia Else on 20 Jun 2010 02:07 On 20/06/2010 3:48 PM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... >> On 20/06/2010 6:10 AM, |-|ercules wrote: >> >>> If all digits of a single infinite expansion can be contained with >>> increasing finite prefixes, >>> and the computable set of reals has EVERY finite prefix, then all digits >>> of EVERY infinite >>> expansion are contained. >> >> It's far from clear what that actually means, but in any case you >> certainly haven't proved it. >> >> Sylvia. > > You agreed that all digits of PI (in order) are contained in this list, > right? > > 3 > 31 > 314 > ... > > So you should get the first part, > >>> If all digits of a single infinite expansion can be contained with >>> increasing finite prefixes, > > Herc I've told you which step I have reservations about. Sylvia.
From: K_h on 20 Jun 2010 03:20 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4c1d7dce$0$316$afc38c87(a)news.optusnet.com.au... > >>> My whole argument is that they cannot be listed in their >>> entirety, or we could use a Cantor construction to >>> produce a computable Real not on the list. >> >> You are wrongly assuming that only computable sets exist. > > No. Good. Then why wouldn't the "Cantor construction" you talk about above produce a non-computable real? _
From: WM on 20 Jun 2010 04:27 On 20 Jun., 01:58, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-19, Virgil <Vir...(a)home.esc> wrote: > > > There is, however, some question in my mind about the existence of a > > list of all and ONLY computable reals. > > Why? The computable reals can be proven countable, as you already > know, so there is a bijection between N and the set of countable > reals. That is not true. An exclusive list need not exist. The reals in a certain Cantor-list are countable. And if you form the anti-diagonal of that list and add it (for instance at first position) to the list, this new list is also countable. Again consttruct the anti-diagonal and add it to the list. Continue. The situation remains the same for all anti-diagonals you might want to construct. Therefore all reals constructed in that way are countable. Nevertheless it is impossible to put all of them in one list, because there would be another resulting anti-diagonal. Conclusion: It is impossible to obtain a bijection of all these reals with N although all "these" reals are countable with no doubt. Regards, WM
From: WM on 20 Jun 2010 04:31
On 20 Jun., 02:04, "Mike Terry" > > No, you're misunderstanding the meaning of computable. > > Hopefully you will be OK with the following definition: > > A real number r is computable if there is a TM (Turing machine) > T which given n as input, will produce as output > the n'th digit of r. Whatever might be the true meaning: The Turing machine need a finite definition. Therefore the computable number has a finite definition. There are only countable many finite definitions. And every diagonal of a defined Cantor list has also a finite definition. Therefore Cantor shows that the countable set of all real numbers with finite definitions is uncountable (or that the defined diagonal number is undefined). Regards, WM |