THE CANTOR ARGUMENT SO FAR
in [Math]
|
Prev: JSH: Judging difficulty
Next: HOW DID GOVERNMENT BEGIN?
From: Sylvia Else on 20 Jun 2010 22:29 On 21/06/2010 11:48 AM, Mike Terry wrote: > "Sylvia Else"<sylvia(a)not.here.invalid> wrote in message > news:887pi5F9g6U1(a)mid.individual.net... >> 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. > > This isn't right. If you could do that, you could arrange for the > antidiagonal of the permuted list to be in the list, which would be a > contradiction. (So you can't do it!) I don't see that. If a number isn't in a list, then it's not in any permutation of the list either. > > Of course by permuting the list before taking the antidiagonal you can > create a large supply of different numbers which are ALL not in the original > list, but there ARE constraints on the numbers you can produce this way. Well, it's true that I rather assumed, without proof, that any number can be created in the diagonal. But I don't think anything relevant turns on that. > > As you say, none of this matters for Cantors proof, as it only needs ONE > such number - producing 100 missing numbers could be called "misplaced > mathematical enthusiasm". > Herc certainly seems to have that :) Sylvia.
From: Sylvia Else on 21 Jun 2010 00:12 On 21/06/2010 1:14 PM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote >> Well, it's true that I rather assumed, without proof, that any number >> can be created in the diagonal. But I don't think anything relevant >> turns on that. >> > > You admit your error to him but not me? I haven't admitted that it's not true, only that I can't prove it. However, since it's not relevant anyway, it doesn't matter that I can't prove it. Maybe someone can prove it, or disprove it, and I'd be interested either way. But it doesn't impact on the discussion about Cantor. I disagree with you. That's not the same. > What are we married or something! I hope not. Pretty sure not. > > Atleast you dismissed your error as irrelevant in true womanly fashion. Some errors are relevant, some are not. This one wasn't. Sylvia.
From: Sylvia Else on 21 Jun 2010 01:22 On 21/06/2010 11:48 AM, Mike Terry wrote: > "Sylvia Else"<sylvia(a)not.here.invalid> wrote in message >> 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. > > This isn't right. If you could do that, you could arrange for the > antidiagonal of the permuted list to be in the list, which would be a > contradiction. (So you can't do it!) As I said in my other response getting a number onto the diagonal doesn't mean it's in the list. But that doesn't mean it's possible either. However.... If I require digit n to be d, and assuming it isn't already, I can search down the list until I find a line with digit n equal to d, and then permute the list so that that line is line n. But can I necessarily find such a line? Suppose line n has k as digit n. Then I could get d there instead by adding a constant. Since line n is a computable, the value of that computable plus the constant is also a computable, so must be in the list. But it might be in the preceding n - 1 lines that I cannot touch. However, there are only n - 1 lines, but 10^(n - 1) - 1 different constants I can choose, so I must be able to find a constant to add such that the resulting computable doesn't match any of the preceding n lines. That computable must be somewhere later in the list from where I can permute it. So it seems to me that I can always permute the list to obtain a desired digit, and thus construct any real on the diagonal by permuting a list of computables. Since this means the set of permutations must be uncountably infinite, if follows that there are uncountably many permutations of a countably infinite set. Does that stand up? Sylvia.
From: |-|ercules on 21 Jun 2010 01:44 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 21/06/2010 1:14 PM, |-|ercules wrote: >> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>> Well, it's true that I rather assumed, without proof, that any number >>> can be created in the diagonal. But I don't think anything relevant >>> turns on that. >>> >> >> You admit your error to him but not me? > > I haven't admitted that it's not true, only that I can't prove it. > However, since it's not relevant anyway, it doesn't matter that I can't > prove it. Maybe someone can prove it, or disprove it, and I'd be > interested either way. But it doesn't impact on the discussion about Cantor. It's trivially false as I already said. If one listed real is 0.111... then the diagonal cannot be 0.222... > > I disagree with you. That's not the same. > >> What are we married or something! > > I hope not. Pretty sure not. > >> >> Atleast you dismissed your error as irrelevant in true womanly fashion. > > Some errors are relevant, some are not. This one wasn't. What? If your statement was true, then the fact the diagonal could be anything would trivially disprove that the anti-diagonal was even relevant to the list. Herc
From: Sylvia Else on 21 Jun 2010 02:05
On 21/06/2010 3:44 PM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote >> On 21/06/2010 1:14 PM, |-|ercules wrote: >>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>>> Well, it's true that I rather assumed, without proof, that any number >>>> can be created in the diagonal. But I don't think anything relevant >>>> turns on that. >>>> >>> >>> You admit your error to him but not me? >> >> I haven't admitted that it's not true, only that I can't prove it. >> However, since it's not relevant anyway, it doesn't matter that I >> can't prove it. Maybe someone can prove it, or disprove it, and I'd be >> interested either way. But it doesn't impact on the discussion about >> Cantor. > > It's trivially false as I already said. If one listed real is 0.111... > then the diagonal cannot > be 0.222.. Infinities are tricky things. I can construct an arbitrarily long number that doesn't contain a 1. > > > >> >> I disagree with you. That's not the same. >> >>> What are we married or something! >> >> I hope not. Pretty sure not. >> >>> >>> Atleast you dismissed your error as irrelevant in true womanly fashion. >> >> Some errors are relevant, some are not. This one wasn't. > > > What? If your statement was true, then the fact the diagonal could be > anything would trivially disprove that the anti-diagonal was even relevant > to the list. I don't see how. Sylvia. |