Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 17 Jun 2010 04:05 On 17 Jun., 04:37, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > news:ad6b33fa-7e91-4f0e-a390-30ee1050284b(a)s9g2000yqd.googlegroups.com... > > > > > > > On 16 Jun., 13:15, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > > wrote: > >> "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > >>news:91a57f95-54ee-4f0a-8341-b2a7dc2f11de(a)h13g2000yqm.googlegroups.com... > >> On 16 Jun., 05:37, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > >> > Virgil <Vir...(a)home.esc> writes: > >> > > But until you can determine which of those 10 cases, how can you > >> > > compute the number? > > >> > You can't. The number is computable nonetheless, in the sense that > >> > there > >> > exists an effective procedure for churning out its decimal expansion. > > >> > As noted, computability is a purely extensional notion. Recall the > >> > classical recursion theory exercise, which we find, in some form or > >> > other, in pretty much any text on the subject: > > >> > Let f : N --> N be a function such that > > >> > f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > >> > Is f computable? > > >> All computable, Turing-computable, nonetheless computable, definable > >> (in any useful, i.e., finite language), and by other means > >> determinable numbers form a countable set. > > >> ________________________ > >> All subsets of N are countable. Big deal. > > > All subsets of N that can be specified somehow belong to a countable > > set. > > As I said, all subsets of N are countable. Also as I said, big deal. No. You misunderstood. Please read carefully. The set of all subsets of N that can be specified somehow is countable. > > No, try to think in a rational way. Every real number that can be > > identified by definition, computation, and whatever notions else may > > be invented by set theorists to shore up their broken walls, all these > > numbers belong to a countable set. > > Yes. And so do all subsets of N that can be specified. > > > Therefore Cantor "proves" the uncountability of a countable set. > > No. Cantor's proof is not about a "set" of Reals; it is about a "list" of > Reals. But from this list he concludes that the set of reals that can appear as a diagonal, hence can be specified somehow, is uncountable. This conclusion is wrong > > You seem to not understand that something can be a set but cannot be > enumerated as a list, which is pretty much the whole point of Cantor's > proof. I understand that aleph_0 is considered to be a number. Then we can conclude that even sets that cannot be enumerated, can have cardinal number aleph_0. Because they are a subset of a set that can be enumerated. > > > > >> ____________________ > >> No, it proves there is a Real which is not on the list. > > > What is the interpretation? The reals that can be constructed, are > > uncountable. > > No. The interpretation is that you cannot form a list of all Reals, and any > list which purports to be so must miss at least one Real. The following list contains all reals (i.e. their finite names) that can be specified. This contradicts your claim. 1.1.a 0 2.1.a 1 3.1.a 00 .... Every line can be expanded to a (countable) infinity of an infinity of lines to cover every word in every language and every alphabet. There is no name missing. The list is counatble. > > > Another proof shows: The reals that can be constructed, are countable. > > So there is a contradiction. What is the reason: The assumption of > > finished infinity. > > No, the pretty obvious (and correct) conclusion is that you cannot form a > list of the constructable Reals. > > Note, again, that you are implicitly treating a list of Reals as the same > thing as a set of Reals; Cantor's proof is about Lists, not Sets. Cantor concludes about sets. That is wrong. Further there is a list given above that contains all constructable reals - though some more names. But it would be wrong to assume that a subset has larger cardinal number than its superset. > > > There are no infinite sequences of digits other than established by > > finite formulas as a_n = 1/n or pi or sqrt(2). > > Cantor's list with infinitely many infinitely long lines containing > > arbitrary sequences of digits does not exist (other than by finite > > formulas - but they are countable). In fact nobody has ever seen that > > chimera. A typical case of mass hysteria. How could mathematicians be > > taken in by that rubbish? > > Well, mathematicians don't confuse lists of Reals with sets of Reals. Why then do they claim that the set of reals is uncountable and conclude that 2^aleph_0 is larger than aleph_0? In fact, that is the error. Regards, WM
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 |