CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: Sylvia Else on 26 Jun 2010 23:06 On 27/06/2010 12:56 PM, |-|ercules wrote: > "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. I'm not even sure what that means. Herc, you're the person presenting the proof. You have to prove each step. You can't just assert things and expect people to accept them. > > >> >> "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. If you meant impossible, then the problem with 2 is that you haven't proved it. Sylvia.
From: Sylvia Else on 26 Jun 2010 23:14 On 27/06/2010 1:01 PM, |-|ercules wrote: > "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. Any more than repeating that you've proved it 20 times does. You're presenting the proof. The onus is on you to prove it. > > You said that 10 times already before you admitted you didn't understand > any of > steps 1, 2, 3 and 4. I still don't understand the role of 1. I don't think 2 was ever in doubt as far as it went. 3 was stating the obvious, and we never reached 4. Your proof still seems to depend on showing that the set of all permutations of digits of unbounded but finite length contains all infinite sequences. Your're no closer to proving that than you ever were. Sylvia.
From: |-|ercules on 26 Jun 2010 23:20 "Sylvia Else" <sylvia(a)not.here.invalid> wrote... > On 27/06/2010 12:56 PM, |-|ercules wrote: >> "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. > > I'm not even sure what that means. > > Herc, you're the person presenting the proof. You have to prove each > step. You can't just assert things and expect people to accept them. > >> >> >>> >>> "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. > > If you meant impossible, then the problem with 2 is that you haven't > proved it. I just told you in the other thread, your opinion is moot. You aren't qualified and you have demonstrated confusion over numerous points that everyone else already knows. Stop trolling my proof thread with "THIS ISN'T PROOF" dozens of times. If you think my induction only works on finite prefixes and not over entire infinite expansions then YOU prove that assertion. If you have no other clear objection then shush. (I am aware of the diagonal proof). Herc
From: |-|ercules on 26 Jun 2010 23:22 "Sylvia Else" <sylvia(a)not.here.invalid> wrote in > > You're presenting the proof. The onus is on you to prove it. > That is circular, completely contrary to what a proof is. Herc
From: |-|ercules on 26 Jun 2010 23:23
"Sylvia Else" <sylvia(a)not.here.invalid> wrote >> >> Repeating I haven't proved anything 20 times doesn't make it so. > > Any more than repeating that you've proved it 20 times does. FFS. I'm answering your constant idiotic retorts based on zero qualifications in the field. Herc |