A BLATENT FLAW in Cantor's diag proof
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From: William Hughes on 9 Jun 2010 02:50 On Jun 9, 12:13 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > "William Hughes" <wpihug...(a)hotmail.com> wrote > > > > > On Jun 8, 11:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> "William Hughes" <wpihug...(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? > > > Yes. However, it is not in the list. The list contains > > > 3 > > 31 > > 314 > > 3145 > > ... > > > It does not contain pi. > > > - William Hughes > > Your argument seems to be there is uncountable infinity because computers can only calculate a finite number of digits. No, I have never mentioned computers and the fact that computers can only calculate a finite number of digits is entirely irrelevant. - William Hughes
From: |-|ercules on 9 Jun 2010 02:52 "William Hughes" <wpihughes(a)hotmail.com> wrote > On Jun 9, 3:30 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> >> > On Jun 9, 12:42 am, George Greene <gree...(a)email.unc.edu> wrote: >> >> > "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> > > 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. >> >> >> I said it first. >> >> >> Herc replied (astoundingly) >> >> >> > No. >> >> >> If you actually believe this, then why do you keep talking about how >> >> having "every digit sequence" MATTERS? THIS LIST HAS EVERY >> >> digit sequence, up to EVERY finite length, MATCHING .33333.... ! >> >> NAME ME ONE POSITION where this list of FINITE strings DOESN'T MATCH >> >> .3333.....! YOU CAN'T!! Yet DESPITE this, .3333.... IS NOT ON THIS >> >> LIST! >> >> YOU YOURSELF JUST SAID SO! >> >> > Indeed, and you agreed. There existing some >> > position where the initial part of X >> > does not match any of the Y's is not the only way that >> > X can differ from every one of the Y's . >> >> >> So why are you having so much trouble noticing that EVEN if you have >> >> EVERY possible FINITE initial sequence somewhere on your list of >> >> computable reals, you still DON'T have many infinite ones (and you >> >> provably do NOT >> >> have the infinite anti-diagonal, since that PROVABLY DIFFERS from >> >> EVERYthing you DO have >> >> on the list!)? >> >> > I am not having any trouble. The infinite anti-diagonal >> > is a "new string". It is a string missed by the list. >> > The list does not contain every possible string. >> > The fact that >> >> > All possible digit sequences are computable >> > to all, as in an infinite amount of, >> > finite lengths. >> >> > does not mean there is a list where Cantor's >> > argument fails. >> >> > - William Hughes >> >> There are 2 cases to consider. >> >> 1/ Every finite substring of a certain digit sequence is on some list >> >> 2/ Every finite substring of every possible digit sequence is on some list >> >> 1 does not imply the certain digit sequence is present. >> >> 2 does imply that every digit sequence is present >> > > Nope 2 is true, but the list does not contain a sequence without > last digit. Any sequence without last digit is not on the > list. > > - William Hughes uhuh. I already corrected that, but you ignored the question afterwards which still stands. Are you saying Cantor's diag proof finds an ACTUAL NEW DIGIT SEQUENCE? Even though 2 is true: every possible digit sequence to all (an infinite amount of) finite digits is on the list? Note: there are infinitely many computable reals with infinite expansions, i.e. no last digit but they STILL cover every possible digit sequence. Herc
From: |-|ercules on 9 Jun 2010 02:57 "William Hughes" <wpihughes(a)hotmail.com> wrote > On Jun 9, 12:13 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> >> > On Jun 8, 11:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> "William Hughes" <wpihug...(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? >> >> > Yes. However, it is not in the list. The list contains >> >> > 3 >> > 31 >> > 314 >> > 3145 >> > ... >> >> > It does not contain pi. >> >> > - William Hughes >> >> Your argument seems to be there is uncountable infinity because computers can only calculate a finite number of digits. > > No, I have never mentioned computers and the fact that computers > can only calculate a finite number of digits > is entirely irrelevant. > - William Hughes Wasn't it you who said the computer would never get to the second real if the first was 00000... Herc
From: William Hughes on 9 Jun 2010 02:59 On Jun 9, 3:52 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > "William Hughes" <wpihug...(a)hotmail.com> wrote > > > > > On Jun 9, 3:30 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> "William Hughes" <wpihug...(a)hotmail.com> wrote > > >> > On Jun 9, 12:42 am, George Greene <gree...(a)email.unc.edu> wrote: > >> >> > "William Hughes" <wpihug...(a)hotmail.com> wrote > >> >> > > 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. > > >> >> I said it first. > > >> >> Herc replied (astoundingly) > > >> >> > No. > > >> >> If you actually believe this, then why do you keep talking about how > >> >> having "every digit sequence" MATTERS? THIS LIST HAS EVERY > >> >> digit sequence, up to EVERY finite length, MATCHING .33333.... ! > >> >> NAME ME ONE POSITION where this list of FINITE strings DOESN'T MATCH > >> >> .3333.....! YOU CAN'T!! Yet DESPITE this, .3333.... IS NOT ON THIS > >> >> LIST! > >> >> YOU YOURSELF JUST SAID SO! > > >> > Indeed, and you agreed. There existing some > >> > position where the initial part of X > >> > does not match any of the Y's is not the only way that > >> > X can differ from every one of the Y's . > > >> >> So why are you having so much trouble noticing that EVEN if you have > >> >> EVERY possible FINITE initial sequence somewhere on your list of > >> >> computable reals, you still DON'T have many infinite ones (and you > >> >> provably do NOT > >> >> have the infinite anti-diagonal, since that PROVABLY DIFFERS from > >> >> EVERYthing you DO have > >> >> on the list!)? > > >> > I am not having any trouble. The infinite anti-diagonal > >> > is a "new string". It is a string missed by the list. > >> > The list does not contain every possible string. > >> > The fact that > > >> > All possible digit sequences are computable > >> > to all, as in an infinite amount of, > >> > finite lengths. > > >> > does not mean there is a list where Cantor's > >> > argument fails. > > >> > - William Hughes > > >> There are 2 cases to consider. > > >> 1/ Every finite substring of a certain digit sequence is on some list > > >> 2/ Every finite substring of every possible digit sequence is on some list > > >> 1 does not imply the certain digit sequence is present. > > >> 2 does imply that every digit sequence is present > > > Nope 2 is true, but the list does not contain a sequence without > > last digit. Any sequence without last digit is not on the > > list. > > > - William Hughes > > uhuh. I already corrected that, but you ignored the question afterwards which still stands. > > Are you saying Cantor's diag proof finds an ACTUAL NEW DIGIT SEQUENCE? > Even though 2 is true: every possible digit sequence to all (an infinite amount of) finite > digits is on the list? Yes, every sequence on the list has a last digit. Cantor's diagonal proof finds a sequence without a last digit. Since this sequence is not in the list it is an "actual new digit sequence" > Note: there are infinitely many computable reals with infinite expansions, i.e. no last digit > but they STILL cover every possible digit sequence. Nope. They cover every possible digit sequence with a last digit. They do not cover every possible digit sequence. - William Hughes
From: |-|ercules on 9 Jun 2010 03:24
"William Hughes" <wpihughes(a)hotmail.com> wrote > Yes, every sequence on the list has a last digit. >> Note: there are infinitely many computable reals with infinite expansions, i.e. no last digit I don't see your argument as you seem to be disputing my note. Herc |