Why can no one in sci.math understand my simple point?
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From: WM on 16 Jun 2010 05:03 On 15 Jun., 22:45, Virgil <Vir...(a)home.esc> wrote: > In article > <f78b53d6-24d1-42e2-86bd-1dd0893b8...(a)q12g2000yqj.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > 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. > > WM would prohibit others from doing precisely what he does himself so > often? > > > 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. > > If that list is .0, .1, .00, ..., then it contains no naturals greater > than 1. This list is the list of all words possible in any language based upon any finite alphabet. The list is given in binary. All alphabets, all languages and all dictionaries are contained in later, rather long but finite lines. Therefore this is a list of everything (that can meaningfully be expressed). This list does not allow for a diagonal, because that is a meaningless concept. (That is proved in my list, in a later line.) Regards, WM
From: WM on 16 Jun 2010 05:06 On 15 Jun., 22:52, Virgil <Vir...(a)home.esc> wrote: > In article > <4b892c9b-5125-46b6-8136-4178f0aca...(a)b35g2000yqi.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > 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. > > There are undecidable propositions in mathematics, so if P is one of > them then "x = 1 if P is true otherwise x = 0" defines an uncomputable > number. 3 is a number. n is a number form. 5 < 7 is an expression. m < n is the form of an expression. f(n) = (1 if Goldbach is correct) is not a function and it is not computable. It is a form of a function or of a sequence. Regards, WM
From: WM on 16 Jun 2010 05:11 On 16 Jun., 00:13, "K_h" <KHol...(a)SX729.com> wrote: > > No one item on the list contains pi in its entirety. > > True, there is no entry for pi, in its entirety, on the list > but all of the digits of pi are there along the diagonal. By induction we prove: There is no initial segment of the diagonal that is not as a line in the list. And there is no part of the diagonal that is not in one single line of the list. Regards, WM
From: |-|ercules on 16 Jun 2010 05:48 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote ... > > By induction we prove: There is no initial segment of the (ANTI)diagonal > that is not as a line in the list. Right, therefore the anti-diagonal does not contain any pattern of digits that are not computable. Cantor's trick 123 456 789 Diag = 159 AntiDiag = 260 never produces a new sequence of digits. Herc
From: WM on 16 Jun 2010 07:02
On 16 Jun., 02:39, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > 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. > > Yes. If you provide a list of Reals, then the diagonal is computable and > does not appear on the list. Delicious. Cantor shows that the countable set of computable reals is uncountable. > > > Then the proof shows > > that the countable set of defined reals is uncountable. > > No, it shows that all "definable" (computable) Reals cannot be explicitly > listed. This is *not* the same as being uncountable. It is impossible to explicitly list anything infinite. You can *define* a list by a_n = 1/n. But that is not the same as explicitly list 1/1, 1/2, 1/3, and so on. > > > Or it does not > > show anything at all. > > It shows that all "definable" (computable) Reals cannot be explicitly > listed. This is a well known proof in set theory. This is *not* the same as > being uncountable. Cantor "proved" that the reals are uncountable because they cannot be listed. Now you say: Unlistability has nothing to do with uncountability. > > > 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. > > No, that paragraph above is close to gibberish. Cantor said and proved that > any purported list of all Reals cannot contain all Reals. His proof is > simple and clear, provides an explicit construction of at least one missing > Real, and does contain or require any concepts of uncomputable numbers, or > use of the Axiom of Choice. The proof is invalid. What is proved is: Any finite initial segment of a list does not contain the finite diagonal number. That is correct. It is impossible at all to obtain a number from an infinite sequence because the number cannot be known unless the sequence has been finished. But an infinite sequence is never finished. For example: You cannot find out what number I have in mind when writing 0.1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111and so on. You may guess that I mean 1/9 but I could also intend to write a 2 at a later position or to switch to zeros. I am deeply impressed how few people see that from a finite definition D one can obtain an infinite sequence S: D ==> S but that the arrow cannot be reversed: S ==> D is wrong unless the S is completely given. It is never complete, however. > > Perhaps if you were to identify the step in Cantor's proof that you consider > wrong, then we might gain some idea as to what you are actually objecting > to? Of course. Cantor assumes that the list represents finished infinity. But infinity cannot be finished by definition. Without finished infinity, there is always a further line of the list. Regards, WM |