THE CANTOR ARGUMENT SO FAR
in [Logic]
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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.
From: |-|ercules on 21 Jun 2010 02:23
"Sylvia Else" <sylvia(a)not.here.invalid> wrote > 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. huh? > >> >> >> >>> >>> 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. I think your brain serum run out. Herc |