CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: Sylvia Else on 26 Jun 2010 23:44 On 27/06/2010 1:20 PM, |-|ercules wrote: > "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. I don't think that's actually how proofs work. I've already shown that your induction doesn't produce a list of all infinite sequences, because 1/9 isn't in the list - it would have infinitely many predecessors. Sylvia.
From: Sylvia Else on 26 Jun 2010 23:46 On 27/06/2010 1:22 PM, |-|ercules wrote: > "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. > Perhaps we have different notions about what constitutes a proof. Are you under the impression that a proof consists of a set of assertions that no one has challenged, (or in your case, that no one whose qualifications you accept has challenged)? Sylvia.
From: Tim Little on 27 Jun 2010 00:01 On 2010-06-27, Sylvia Else <sylvia(a)not.here.invalid> wrote: > I've already shown that your induction doesn't produce a list of all > infinite sequences, because 1/9 isn't in the list - it would have > infinitely many predecessors. Yes, and what's worse it doesn't even have a direct predecessor. Induction starts with a base case "P(0) is true" and uses implication "P(n)->P(n+1)" to prove "for all natural numbers n, P(n) is true". There is no sequence length "n" for which an infinite sequence appears in length "n+1", so induction doesn't apply. - Tim
From: |-|ercules on 27 Jun 2010 00:15 "Tim Little" <tim(a)little-possums.net> wrote > On 2010-06-27, Sylvia Else <sylvia(a)not.here.invalid> wrote: >> I've already shown that your induction doesn't produce a list of all >> infinite sequences, because 1/9 isn't in the list - it would have >> infinitely many predecessors. huh? this is total garbage, the Complete Permutation Sets are sets of oo long reals. I've explained this 5 times already. It is the prefixes of complete permutations that are 'finite', not the reals themselves. > > Yes, and what's worse it doesn't even have a direct predecessor. > Induction starts with a base case "P(0) is true" and uses implication > "P(n)->P(n+1)" to prove "for all natural numbers n, P(n) is true". > There is no sequence length "n" for which an infinite sequence appears > in length "n+1", so induction doesn't apply. The proof is so simple I am surprised nobody can spot the inductive step. The initial step is a trivial case of all finite prefixes of digit width 1, e.g. CPS_1 = { 0.04748574849.. , 0.14848498483.. , 0.24848439494.. , 0.34505940940.. , 0.45059594040.. , 0.54584848483.. , 0.60606060606.. , 0.73748747484.. , 0.83834838484.. , 0.94834748484.. , } Back to Sylvia, you just made a lucid post that my proof COULD work, and the attacks based on ZFC are assumptive. This contradicts your other claims that the proof is not valid. When you say "YOU HAVEN'T PROVEN THAT" you should word it "YOU HAVEN'T PROVEN THAT TO ME". I allow that statement. Herc
From: George Greene on 27 Jun 2010 00:28
On Jun 26, 3:06 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Isn't this just permutation with replacement? No. You can't replace anything on any of the lists we are talking about. Everything on them is a constant, not a variable. And "with replacement" IS NOT EVEN A KIND of permutation! GOOD GRIEF! "With replacement" is something you use when pulling stones out of a jug or something. THIS IS NOTHING LIKE THAT. ALL of this stuff IS FIXED AND CONSTANT. |