THE CANTOR ARGUMENT SO FAR
in [Logic]
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From: |-|ercules on 20 Jun 2010 12:40 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 20/06/2010 7:42 PM, |-|ercules wrote: >> "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... >>> On 20/06/2010 5:03 PM, |-|ercules wrote: >>>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... >>>>> On 20/06/2010 12:42 PM, |-|ercules wrote: >>>>>> "George Greene" <greeneg(a)email.unc.edu> wrote >>>>>>>> Is it a *NEW DIGIT SEQUENCE* or not? >>>>>>> >>>>>>> YES, DUMBASS, IT IS A NEW DIGIT-SEQUENCE because it was >>>>>>> NOT ON THE LIST of (allegedly "all") THE OLD digit-sequences! >>>>>> >>>>>> >>>>>> But you keep saying the anti-diagonal is NEW and ignoring me when >>>>>> I say it's not a new digit sequence. >>>>>> >>>>>> Then you repeat Cantor's proof again and again that it's NEW. >>>>>> >>>>>> You use terms NEW and NOT ON THE LIST, but evade me when I challenge >>>>>> whether it contains any new digit sequence. >>>>> >>>>> And you just ignore the point that it must be new because it >>>>> demonstrably isn't in the list. >>>>> >>>>> Sylvia. >>>> >>>> All you demonstrated was >>> >>> In what sense could a number be said to be in a list if it doesn't >>> match any element of the list? >> >> >> What sense is a number that's only definition is to not be on a list? >> >> That may seem less concrete than your question, but it's worth thinking >> what "anti-diagonals" entail. >> >> You're not just constructing 0.444454445544444445444.. a 4 for every non >> 4 digit and a 5 for a 4. >> >> You're constructing ALL 9 OTHER DIGITS to the diagonal digits. > > Why? We only need one number that's not in the list. Doing what you > suggest just creates many more numbers that are not in the list. > >> >> And it's not just the diagonal, it's the diagonal of ALL PERMUTATIONS OF >> THE LIST. > >> >> So, the first digit of the list can be... well anything, so the >> antidiagonal starts with anything.. >> then the second digit of the second real can be anything, so the >> antidiagonals next digit is anything.. > > The problem with that is before you can permute a list, you have to > prove that a list exists. If it doesn't exist, you can't permuted it. > Cantor assumes that a list exists, and then proceeds to show that that > leads to a contradiction. Which is fair enough. But you can't just > assume the list exists, and then use arguments about permuting the list > to prove that it exists, or to negate the proof that it doesn't exist. > That's circular. > > <snipped rest based on false premise> > > Sylvia. So you can form an anti-diagonal on a hypothetical list like the computable reals but you can't reorder that list? If you take the 2nd digit of the 2nd real, then the 1st digit of the 1st real, then the remaining nth digit of the nth reals, the remainder of the usual diagonal, that won't work? 0. _ x _ _ _ _ _ 0. x _ _ _ _ _ _ 0. _ _ x _ _ _ _ 0. _ _ _ x _ _ _ 0. _ _ _ _ x _ _ 0. _ _ _ _ _ x _ 0. _ _ _ _ _ _ x Does not an anti-diagonal make! Herc
From: |-|ercules on 20 Jun 2010 12:41 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 20, 3:03 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> All you demonstrated was >> >> An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) > > > This is NOT ALL we demonstrated. > BECAUSE we demonstrated this, we HAVE THEREFORE demonstrated that > AD(L) IS NOT ON the list L! > >> You don't like axioms stating a fact, > > YOU ARE A DAMN LIAR AND A DAMN FOOL. > WE DO SO TOO like "Axioms stating a fact"! That's what an axiom IS! > It is A FACT about the little abstract/mathematical sub-universe that > WE are TRYING > to talk about (we call this "the universe of discourse", for this > PARTICULAR discourse -- > if you want to have a DIFFERENT discourse about OTHER things, FINE). > >> but you use a definition as a proof. > > Dipshit: YOU DON'T KNOW THE DIFFERENCE between a definition and a > proof. > JUST WHAT definition did we use? > And we don't use anything ELSE AS a proof: > we use A PROOF as a proof! > Cantor's Theorem HAS A PROOF (that's what MAKES it a theorem) > and YOU HAVE SEEN this proof! > And you have NOT shown any errors in it! > Yet you persist! > I was referring to Sylvia's comment: > Hardly. I'm just saying that it's uninteresting that a statement is true > under an axiom when it's just a restatement of the axiom. Herc
From: |-|ercules on 20 Jun 2010 12:48 (typo) > If you take the 1st digit of the 2nd real, then the 2nd digit of the 1st real, then > the remaining nth digit of the nth reals, the remainder of the usual diagonal, that won't work? > > 0. _ x _ _ _ _ _ > 0. x _ _ _ _ _ _ > 0. _ _ x _ _ _ _ > 0. _ _ _ x _ _ _ > 0. _ _ _ _ x _ _ > 0. _ _ _ _ _ x _ > 0. _ _ _ _ _ _ x > > Does not an anti-diagonal make! > > Herc
From: |-|ercules on 20 Jun 2010 12:52 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 20, 2:59 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> I still maintain all possible variations of digit sequences are present >> up to infinite width on the list. > WHAT list? Computable reals. Herc
From: George Greene on 20 Jun 2010 15:26
On Jun 20, 12:41 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > Hardly. I'm just saying that it's uninteresting that a statement is true > > under an axiom when it's just a restatement of the axiom. That the anti-diagonal is not on the list is NOT re-statement of any axiom. |