CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: George Greene on 26 Jun 2010 13:19 On Jun 25, 10:08 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > So now what? I myself have nothing against induction schemata > like those of Herc DAMN, you're STUPID. In the first place, Herc DOES NOT HAVE an INDUCTION SCHEMA! He has quantifier dyslexia and an illegitimate substitution of "infinity" as the peg in the HOLE where you need "the maximum" of some statistic over some collection (the point being that infinite collections don't have to actually CONTAIN their least upper bound). In the second place, if you have accurately captured, via an induction schema that Herc himself cannot understand, what Herc IS ACTUALLY doing by way of reasoning here, then, in case you hadn't noticed, THAT SCHEMA IS LOGICALLY INCONSISTENT as herein applied! It PRODUCES CONTRADICTIONS! This constitutes an indirect PROOF that one of the premises IS FALSE! Like Herc's schema, maybe, if the rest of your machinery was sound?? So isn't PROVING SOMETHING FALSE very MUCH *having*something*against* it?? You know, something like IT'S FALSE???? I repeat, DAMN, you're STUPID.
From: George Greene on 26 Jun 2010 13:23 On Jun 25, 10:08 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > So now we can see why schemata like Herc's are usually > unpalatable to Herc-"religionists." The induction forces every > list that contains 0.1, 0.11, 0.111, etc. to contain 1/9, including > lists that were specifically constructed to contain 0.1, 0.11, > 0.111, etc. and _not_ 1/9, like List Y above. The schema > forces additional structure onto a list NO, IT DOESN'T, DUMBASS!!!! *JEEZUS*!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Precisely as you said, the list WAS EXPLICITLY CONSTRUCTED *NOT* to contain 1/9 !! The list ONLY has FINITE-width elements! The logical conclusion that it ALSO contains 1/9 is therefore A LOGICAL CONTRADICTION! The list cannot BOTH *fail* to contain 1/9 (as required by its definition and construction) AND *contain* 1/9 (as required by Herc's schema)!! Herc's schema *IS THEREFORE*BULLSHIT*!! It is just WRONG! It is just FALSE! You cannot "force additional structure" onto a list THAT IS ALREADY THERE! You cannot "force additional structure" onto pi, or e, or 2! DON'T YOU*KNOW*THE*DIFFERENCE*BETWEEN a VARIABLE and a CONSTANT????
From: |-|ercules on 26 Jun 2010 15:06 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 26, 6:45 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> There seems to be 2 possibilities. >> >> 1 All finite digit permutations occur to infinite length. > > This IS NOT a possibility because for a FINITE permutation to occur > "to infinite length" IS JUST MEANINGLESS! > "To EVERY FINITE length" DOES NOT MEAN "to infinite length", > yet you CONTINUE TO TALK as though it did! > This has just gone on LONG ENOUGH! > > Here is another point: if you are only talking about FINITE lists of > digits, > then "permutation" is IRRELEVANT because ANY permutation of any > finite list of digits IS JUST ANOTHER FINITE LIST OF DIGITS, that > WAS ALREADY IN the class (of finite lists of digits) that you were > talking about! > Finite SEQUENCES of digits or finite LISTS of digits are what are > actually being > dealt with here. A permutation IS A CHANGE OF ORDER and would require > some > notion of WHAT AN "original" order was for a given finite basket of > digits! > There IS NO such notion for you here! STOP SAYING "permutation"! > YOU DON'T KNOW what a permutation is! Isn't this just permutation with replacement? Herc
From: |-|ercules on 26 Jun 2010 18:05 "George Greene" <greeneg(a)email.unc.edu> wrote >> On 26/06/2010 8:45 PM, |-|ercules wrote: >> > 2 It would be very difficult to come up with a new sequence of digits >> > that wasn't on the computable reals list > > On Jun 26, 8:09 am, Sylvia Else <syl...(a)not.here.invalid> wrote: >> The anti-diagonal wouldn't be on it. Now, you might feel that the >> anti-diagonal is obviously computable, in that each digit can be >> obtained algorithmically from the list. > > Of course we feel that! It IS IMPORTANT that anti-diagonalization (of > any > square list) IS computable! THERE IS a TM that does that! > >> But that presupposed that a list >> of computable reals is itself computable, which you haven't proved. We >> know that it is countable, but that's not the same thing. > > PLEASE! That is NOT the point! The point IS that EVEN IF the list of > computable reals is computable, THE FACT THAT the anti-diagonal > IS THEN ALSO computable *produces*a*CONTRADICTION* because > the anti-diagonal both 1) Must Be On the list, since it is computable, > and > 2) CAN'T BE ON the list, BECAUSE it IS the ANTI-diagonal! > > So what is Herc to conclude FROM THAT?? Your 'IS THEN ALSO computable' is as far as we can comment. It's a hypothetical list so the computability of the diagonal is irrelevant. It is perplexing if outputs of all computer programs are listed, how do you find a program to compute the diagonal digit at the position that is contradictory? Perhaps ALL computable outputs is a heirarchy of possible programs acting on outputs of lower level sets of programs. Herc
From: |-|ercules on 26 Jun 2010 18:06
"Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 26/06/2010 8:45 PM, |-|ercules wrote: >> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>> On 26/06/2010 3:40 AM, Graham Cooper wrote: >>>> On Jun 25, 8:23 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: >>>>> On 25/06/2010 7:07 PM, Graham Cooper wrote: >>>>> >>>>> >>>>> >>>>> >>>>> >>>>>> On Jun 25, 6:54 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: >>>>>>> OK. >>>>> >>>>>>> "1 start with an infinite list of all computable reals". >>>>> >>>>>>> That is any list of all the computable reals, howsoever constructed. >>>>> >>>>>>> "2 let w = the maximum width of complete permutation sets" >>>>> >>>>>>> Where a complete permutation set is all the possible combinations of >>>>>>> some finite number of digits. So this step doesn't involve doing >>>>>>> anything with the list described in step 1? It's a completely >>>>>>> independent step? >>>>> >>>>>>> Sylvia. >>>>> >>>>>> Hmmm. Did you consider that the CPS found in the list of >>>>>> step 1 was what I meant. Step 1 - consider this list... >>>>> >>>>> I'm reluctant to assume you mean anything unless it's stated. It seems >>>>> to cause difficulties. However, apparently CPS is an abbreviation for >>>>> "complete permutation set". >>>>> >>>>> So the list of all computable reals contains as a subset complete >>>>> permutation sets whose width is unbounded. Slightly rewording 2, >>>>> gives us: >>>>> >>>>> "2 let w = the maximum width of those complete permutation sets" >>>>> >>>>> and the next step is >>>>> >>>>> "3 contradict 2" >>>>> >>>>> How is it to be contradicted? >>>>> >>>>> Sylvia. >>>> >>>> There is no (finite) maximum. >>> >>> So there is no finite maximum. How does that advance your proof? >> >> >> There seems to be 2 possibilities. >> >> 1 All finite digit permutations occur to infinite length. > > It's certainly true that any digit permutation of finite length will be > on the list, but that's not a proof that all infinite sequences are on > the list. > > Certainly some infinite sequences are on the list, because they are > computable, but you haven't proved that all infinite sequences are on > the list. > >> >> 2 It would be very difficult to come up with a new sequence of digits >> that wasn't on the computable reals list > > The anti-diagonal wouldn't be on it. Now, you might feel that the > anti-diagonal is obviously computable, in that each digit can be > obtained algorithmically from the list. But that presupposed that a list > of computable reals is itself computable, which you haven't proved. We > know that it is countable, but that's not the same thing. > > Sylvia. You assumed the 'finite sequences only' analogy to my proof to claim 1. Then you used 1 to imply 2, using the result of the diagonal attack to discredit my attack of the diagonal attack. Herc |