A BLATENT FLAW in Cantor's diag proof
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From: Tim Little on 8 Jun 2010 20:10 On 2010-06-08, William Hughes <wpihughes(a)hotmail.com> wrote: > I would prefer to say that the diagonalization argument > shows that there is no list of real numbers. In this > form it is true if you only allow computable numbers > (thus only computable lists) [...] You need to be careful here. A computable list must contain only computable numbers, but not all lists of computable numbers are computable lists. Only finite lists of computable numbers are necessarily computable. Herc made that mistake when he said that he could compute a list of all computable numbers. That is impossible. - Tim
From: Daryl McCullough on 8 Jun 2010 20:22 Here's what's funny about USENET. In a regular classroom, you have one teacher and many students. In a typical USENET discussion, there are many teachers and just one student. You'd think that such a low student/teacher ratio would make for quick progress, but that doesn't turn out to be the case. Herc's ignorance can defeat any number of teachers, no matter how knowledgeable and patient. -- Daryl McCullough Ithaca, NY
From: herbzet on 8 Jun 2010 20:37 Daryl McCullough wrote: > > Here's what's funny about USENET. In a regular classroom, > you have one teacher and many students. In a typical USENET > discussion, there are many teachers and just one student. > You'd think that such a low student/teacher ratio would make > for quick progress, but that doesn't turn out to be the case. > Herc's ignorance can defeat any number of teachers, no matter > how knowledgeable and patient. -- hz Against stupidity, the gods themselves contend in vain. -- Schiller --
From: William Hughes on 8 Jun 2010 20:53 On Jun 8, 9:10 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-08, William Hughes <wpihug...(a)hotmail.com> wrote: > > > I would prefer to say that the diagonalization argument > > shows that there is no list of real numbers. In this > > form it is true if you only allow computable numbers > > (thus only computable lists) [...] > > You need to be careful here. A computable list must contain only > computable numbers, but not all lists of computable numbers are > computable lists. Agreed. The reason that only allowing computable numbers restricts you to computable lists is not that a computable list can only contain computable numbers (though this is true) but if you allow arbitrary sequences of computable numbers you can easily get non-computable numbers (e.g. look at equivalence classes of Cauchy sequences), As you note a special case of particular interest in this context is that the list of all computable numbers is not computable. - William Hughes
From: |-|ercules on 8 Jun 2010 22:21
"William Hughes" <wpihughes(a)hotmail.com> wrote > On Jun 8, 6:17 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> >> > On Jun 8, 5:52 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> > On Jun 8, 5:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> >> > On Jun 8, 5:29 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> >> >> The infinitely many long sequences of all possible digit sequences DOESN'T MISS A SEQUENCE OF DIGITS. >> >> >> >> > Since every sequence has a last digit, any sequence >> >> >> > of digits that does not have a last digit is missed. >> >> >> >> You're as confused as when you said there is no algorithm to produce an infinite list. >> >> >> >> Herc >> >> >> > Is this digit sequence (which does not have a last 3) >> >> >> > 33333... >> >> >> > in this list >> >> >> > 1 3 >> >> > 2 33 >> >> > 3 333 >> >> > ... >> >> >> > of sequences (all of which have a last 3). >> >> >> > Yes or No. >> >> >> > - William Hughes >> >> >> No. >> >> > So a list of sequences with last digit will miss >> > a sequence without last digit. >> >> > - William Hughes >> >> Yes, this is a fine QUANTITATIVE argument that SUPPORTS *missing sequences*, using an example of a converging sequence. >> >> Unfortunately a rudimentary QUALITATIVE analysis contradicts that modifying the diagonal results in a new sequence of digits. > > Nope, The list contains only sequences with last digit. A sequence > without last digit is a sequence that > is not contained in the list. > > - William Hughes > Is pi computable? Herc |