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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.
From: George Greene on 20 Jun 2010 15:27 On Jun 20, 12:59 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Yes I have disproven Cantor, see CANTOR DISPROOF <<<<<< > > I've also proven I'm a deity on 02 This explains a lot. Sorry I was so slow to get it.
From: |-|ercules on 20 Jun 2010 15:36 "George Greene" <greeneg(a)email.unc.edu> wrote -37d5ebb92b93(a)u7g2000yqm.googlegroups.com... > 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. > That's twice running you took statements out of context, you really should try to follow the argument better. Sylvia was talking about MY axiom that reals are computable, hence my comment that you don't like axioms being facts. Herc
From: |-|ercules on 20 Jun 2010 15:37 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 20, 12:59 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> Yes I have disproven Cantor, see CANTOR DISPROOF <<<<<< >> >> I've also proven I'm a deity on 02 > > This explains a lot. > Sorry I was so slow to get it. Shame you ignored the proof presented. Herc
From: Sylvia Else on 20 Jun 2010 20:20
On 21/06/2010 2:40 AM, |-|ercules wrote: > "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 It would also be a number that is not in the list. What difference would that make? As usual, it's unclear where you think you're going with this. OK.... You can't permute a list of reals, because there is no such list. If could construct a list of all computable reals, then you could permute the list, generating as many anti-diagonals as you like, and construct thereby numbers that are not in the list of computable reals. All you would be doing is demonstrating that that there are reals that are not computable. Similarly, you can permute a list of computable reals so that any desired number, computable or otherwise, lies on the diagonal, but the existence of the number on the diagonal doesn't mean that it's computable. Or you can construct a list, finite or otherwise, of a subset of the reals, and then use anti-diagonalisation to contruct humbers that are not in the list. You could then add them to the list, and repeat. You end up with an arbitrary countable subset of the reals, which is totally uninteresting. Sylvia. |