A BLATENT FLAW in Cantor's diag proof
in [Math]
|
Prev: I'm gonna try this one more time CANTOR DISPROOF
Next: Reliability: Treating the Building & the Fault Line As One & the Same
From: |-|ercules on 9 Jun 2010 01:09 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 9/06/2010 1:50 PM, Aatu Koskensilta wrote: >> stevendaryl3016(a)yahoo.com (Daryl McCullough) writes: >> >>> I don't think that's correct. There is no teacher that is >>> capable of explaining anything to Herc. >> >> Yet there is no shortage of people strangely willing to give it a try. >> > > Er, that's an assumption. People who appear to be attempting to explain > something to Herc may in fact be doing something else entirely, such as > trying to understand how Herc's mind works. > > Herc appears to have a diverse set of false beliefs that he expresses on > Usenet. What I've been hoping to get is a handle on is whether all of > these false beliefs are delusions arising from his mental illness, or > whether some of them are just conventional beliefs of falsehoods, of the > kind found often enough even in people considered sane. > > Of course, there may not be a clear line dividing categories of false > beliefs. > > What's disconcerting is how fragile is the brain's grip on reality, such > that some obscure chemical imbalance can destroy it. > > Sylvia. Have you ever successfully contradicted anything I've said? Herc
From: William Hughes on 9 Jun 2010 02:19 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
From: |-|ercules on 9 Jun 2010 02:30 "William Hughes" <wpihughes(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 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? Herc
From: |-|ercules on 9 Jun 2010 02:43 "|-|ercules" <radgray123(a)yahoo.com> wrote ... > 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 I might retract that implication, due a to a listing of all finite sequences where it doesn't hold. Does this mean I made a mistake? Technically I did make the subject A BLATANT FLAW in Cantor's diag proof so it's one of those deity paradoxes where say I made a hole so large I couldn't jump over it! Herc
From: William Hughes on 9 Jun 2010 02:47
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 |