CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
|
From: |-|ercules on 26 Jun 2010 22:25 "|-|ercules" <radgray123(a)yahoo.com> wrote > Then you used 1 to imply 2 should read, then you used 1 to imply not 2. Herc
From: Sylvia Else on 26 Jun 2010 22:45 On 27/06/2010 8:06 AM, |-|ercules wrote: > "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 The anti-diagonal is merely an example of a number that isn't on the list. Leave out my comment about the anti-diagonal, and go back a moment. "1 All finite digit permutations occur to infinite length." What exactly does this mean? It doesn't prove that all infinite sequences are present. "2. It would be very difficult to come up with a new sequence of digits that wasn't on the computable reals list." It might well be very difficult. But so what? Very difficult isn't the same as impossible. Sylvia.
From: Sylvia Else on 26 Jun 2010 22:55 On 27/06/2010 3:01 AM, George Greene 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?? One would conclude that the list of computable reals isn't computable. But it's clear that Herc doesn't accept the anti-diagonal argument. Herc can continue with his proof as long as he can prove the steps. In principle, he might end up proving a theorem that contradicts Cantor, in which case the set of axioms would be shown to be inconsistent. If that is his real ultimate goal, then we must be wary of seeking to undermine his proof by arguments that assume that the axioms are consistent. That said, I don't think Herc is going to achieve that, but stranger things have happened. Sylvia.
From: |-|ercules on 26 Jun 2010 22:56 "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... > On 27/06/2010 8:06 AM, |-|ercules wrote: >> "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 > > The anti-diagonal is merely an example of a number that isn't on the > list. Leave out my comment about the anti-diagonal, and go back a moment. > > "1 All finite digit permutations occur to infinite length." > > What exactly does this mean? It doesn't prove that all infinite > sequences are present. True, this is your claim of the proof. > > "2. It would be very difficult to come up with a new sequence of digits > that wasn't on the computable reals list." > > It might well be very difficult. But so what? Very difficult isn't the > same as impossible. > Your construction technique is very simple, but considering every object referred to that I put into words is misinterpreted wherever possible, every little thing I write has to be 100% explicit with no other possible bindings or connotations to any terms, even if I refer to something in the same sentence as "it", or the previous sentence says 'regarding this' or the thread topic is X, you all interpret everything as Y, every other possible object under the sun is given priority over what I'm referring to, so I from now on I will not write 'very difficult' when I mean impossible. I think I actually have made a mistake, I thought I was talking to people not robots who need single interpretation only explicit comments because their contextual disambiguator circuitry has blown a fuse. Herc
From: |-|ercules on 26 Jun 2010 23:01
"Sylvia Else" <sylvia(a)not.here.invalid> wrote ... >>> computable, but you haven't proved that all infinite sequences are on >>> the list. Repeating I haven't proved anything 20 times doesn't make it so. You said that 10 times already before you admitted you didn't understand any of steps 1, 2, 3 and 4. Herc |