Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 15 Jun 2010 13:34 On 15 Jun., 18:46, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > WM says... > > > > >On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> The proof does not make use of any property of infinite lists. > >> The proof establishes: (If r_n is the list of reals, and > >> d is the antidiagonal) > > >> forall n, d is not equal to r_n > > >As every n is finite, it belongs to a finite initial segment of the > >infinite list. > > I'm not sure what you are saying. The fact is, we can prove > that for every real r_n on the list, d is not equal to r_n. Of course. Every real r_n belongs to a finite initial segment of the list. That does not yield any result about the whole list > That means that d is not on the list. There is no extrapolation > involved. Look here: We can prove for any finite segment {2, 4, 6, ..., 2n} of the ordered set of all positive even numbers that its cardinal number is surpassed by some elements of the set. Nevertheless this appears not be a proof that the cardinal number of the whole set is less than some elements of the set. Why should the extrapolation be more valid for Cantor's list? Regards, WM
From: WM on 15 Jun 2010 13:59 On 15 Jun., 18:49, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > WM says... > > > > > > > > >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. > > I was talking about the list of all *computable* reals. There are > definable reals that are not computable, and Cantor's proof shows > how to define one. No, it does not show that because it is impossible to define a real by an infinite definition. A Cantor-list without a finite definition however is an infinite definition. Sensible definitions are finite. A definition allows you to communicate the defined. That is impossible for a Cantor-list, except the list has been constructed according to a finite definition. But then also the diagonal has a finite definition - in every specific language. So you explanation is nonsense in highest degree. > > You can similarly get a list of all definable reals for a specific > language L. Every real that is definable in a specific language is definable in another specific language. The set does not grow or shrink or change when the language is changed. No, my list contains all words in all languages, even the definitions of the languages and all the dictionaries. There is nothing else. > Then Cantor's proof allows us to come up with a new > real that is not definable in language L. (It is definable in a > new language that extends L). The diagonal procedure does not define new languages. It uses simply the diagonalisation. The language that I use is not to be extended. There a real either has a finite definition or it has not. There does no extension help. When have you obtained the meaning of a real number from an infinite sequence of digits for the first time? Regards, WM
From: WM on 15 Jun 2010 14:09 On 15 Jun., 18:53, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > WM says... > > > > >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. > > That's just not true. It is just true. > For example, we can define a real r as follows: > > r = sum from n=0 to infinity of H(n) 2^{-n} > > where H(n) = 1 if Turing machine number n halts on input n, > H(n) = 0 otherwise. > > That's definable, but it is not computable. Anyhow it is not a definition. It would be more useful to define some number by the legs of a crowd of unicorns touching the ground at a given time. Nevertheless your "definition" belongs to a countable set, hence it is no example to save Cantors "proof". Either all entries of the lines of the list are defined and the diagonal is defined (in the same language) too. Then the proof shows that the countable set of defined reals is uncountable. Or it does not show anything at all. To switch "languages" is the most lame argument one could think of. The diagonal argument does not switch languages. And it cannot be applied at all because the list of all finite defiitions does not contain infinite entries. Those however are required for the diagonal argument. Regards, WM
From: WM on 15 Jun 2010 14:18 On 15 Jun., 18:57, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >We should not use oracles in mathematics. > > On the contrary! Many real numbers in physics are not computable > to infinite precision (for example, the fine structure constant). Numbers are computable. The fine structure constant is a name. It has soem 20 letters. > Yet, we can certainly compute other real numbers *relative* to such > parameters. There are a countable number of parameters and a countable number of relative numbers. > We can easily devise an algorithm to compute the square > of the fine-structure constant, for example. This algorithm will > take as an input an approximation to the fine-structure constant, > and will return an approximation to the square of the fine-structure > constant. > > In this sense, the square of any real number is computable relative > to that real number. The fine structure constant, or what we know about it, is a name. This is physics - and it will be the last to support transfinity. Every real number that can be communicated belongs to a countable set. This set includes all names in all languages, all dictionaries (they come at some later lines) and all objects relative to those. 0 1 00 01 10 11 000 .... It is impossible to diagonalise it. It is possible to diagonalise infinite lists only when they have infinite lines. But never anybody has obtained a meaning from an infinite sequence of digits. Beware: This is not an infinite sequence of digits "0.111..." but a finite definition (from which an infinite sequence can be obtained) that cannot be used as a line in an infinite list to be diagonalized. Regards, WM
From: WM on 15 Jun 2010 14:24
On 15 Jun., 19:27, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 15 Jun., 16:23, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > >> "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> writes: > >> > So (B) is equivalent to the statement "there exists an uncomputable > >> > number". > > >> Right. But why then did you say the number was computable? > > > And in what form does it exist? > > This question seems utterly meaningless. That may be the impression of cranks who prefer believing things rather than knowing them. If someone says that somethings exists, then he should be able to explain what that means. For numbers existence is easily proved by giving the value. For uncomputable numbers the existence-question is justified. Regards, WM |