Why can no one in sci.math understand my simple point?
in [Logic]
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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
From: WM on 16 Jun 2010 07:06 On 16 Jun., 03:05, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > "Virgil" <Vir...(a)home.esc> wrote in message > > news:Virgil-3B2F0B.16291815062010(a)bignews.usenetmonster.com... > > > In article <87sk4ohwbt....(a)dialatheia.truth.invalid>, > > Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > > > Virgil <Vir...(a)home.esc> writes: > > > > > 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. > > > > You need infinitely many unknown places. > > > If the value of some decimal digit of a number depends on the truth of > > an undecidable proposition, can such a number be computable? > > Yes - e.g. imagine just the first digit of the following number depends on > an undecidable proposition: > > 0.x000000000... > > There are only 10 possibilities for the number, and in each case it is > obviously computable... These numbers are computable. What you wrote, however, is not a number but a form of a number. 3 > 1 is a correct proposition. x > 1 is the form of a proposition. But all that is not of interest for the present problem: All definable, computable, and somehow identifiable numbers and forms of numbers are within a countable set. Therefore Cantor proves the uncountability of this countable set. Regards, WM
From: J. Clarke on 16 Jun 2010 08:19 On 6/15/2010 4:33 PM, WM wrote: > On 15 Jun., 21:38, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> WM says... >> >> >> >>> On 15 Jun., 18:53, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >>>> 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 certainly is. It uniquely characterizes a real number, >> so it's a definition. > > It does not. If it would, the number could be computed. > Who defines what Turing machine number n would do? Can you say "circular argument"? It's not a number because it's not computable and that proves that all numbers are computable.
From: J. Clarke on 16 Jun 2010 08:05 On 6/15/2010 11:59 AM, Aatu Koskensilta wrote: > WM<mueckenh(a)rz.fh-augsburg.de> writes: > >> We should not use oracles in mathematics. > > Nonsense. Using orcales we can show for example that the P = NP problem > can't be solved using any technique that relativizes. This is a useful > result. > >> A real is computable or not. My list contains all computable numbers: >> >> 0 >> 1 >> 00 >> ... > > Your list doesn't appear to contain any real at all, just finite binary > sequences. Did someone redefine the real numbers to exclude all numbers that consist only of the digits 0 and 1?
From: WM on 16 Jun 2010 11:16
On 16 Jun., 11:48, "|-|ercules" <radgray...(a)yahoo.com> wrote: > "WM" <mueck...(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. > Sorry, you misquoted me. I wrote: 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. But I have to excuse because I wrote somewhat unclear. The meaning is: 1) Every initial segment of the decimal expansion of pi is in at least one line of your list 3. 3.1 3.14 3.141 .... What we can finde in the diagonal (not the anti-diagonal), namely 3.141 and so on, exactly that can be found in one line. This is obvious by construction of the list. 2) Every part of the diagonal is in at least one line. That means, every part is in one single line, or there are parts that are in different lines but not in one and the same. The latter proposition can be excluded. If there are more than one lines that contain parts of pi, then it can be proved, be induction, that two of them contain the same as one of them. This can be extended to three lines and four lines and so on for every n lines. Hence we prove that all of pi, that is contained in at least one of the finite lines of your list, is contained in one single line. Conclusion: Either the complete diagonal pi does not exist, or it exists also in one and the same single line. As the latter is wrong, so is the former. There does not exist an actually infinite sequence. Actually infinite mean finished infinite. That is nonsense. (The proof has the same status as Cantor's diagonal proof. Also his proof is valid only for all finite n. In a similar way we find above: For all finite n: All of pi that is in the list up to line n, is in a single line of the list.) Regards, WM |