A BLATENT FLAW in Cantor's diag proof
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From: Aatu Koskensilta on 8 Jun 2010 23:43 George Greene <greeneg(a)email.unc.edu> writes: > If you order the list properly then the list will be computable as > well. I suggest you revisit Turing's 1936 paper on computable numbers. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Daryl McCullough on 8 Jun 2010 23:45 |-|ercules says... > >"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... >> Here's what's funny about USENET. In a regular classroom, >> you have one teacher and many students. In a typical USENET >> discussion, there are many teachers and just one student. >> You'd think that such a low student/teacher ratio would make >> for quick progress, but that doesn't turn out to be the case. >> Herc's ignorance can defeat any number of teachers, no matter >> how knowledgeable and patient. >If one of the 'teachers' would just answer my questions instead >of putting their fingers in their ears and reading maths scripture >then maybe we could work together on a what Cantor's proofs entail. Some questions indicate that the student lacks certain pre-requisites. In particular, some questions are incoherent. The highest priority in that case is not to answer the question (an incoherent question cannot have any coherent answer) but to get the student to ask a *better* question, one that *can* be given a coherent answer. >TRUE or FALSE > >1/ no box of the box numbers not in their own boxes proves higher >infinities The reason that's an incoherent question is because facts don't prove things. A proof is a *demonstration* that a claim is true. Your thought experiment with boxes is an incoherent mess, but it can be fixed with a little work. Let's try to fix it. You want to say that 1. There is a collection of boxes. 2. Each box contains zero, one or more naturals. 3. Each box is labeled with a natural number. Okay, let's define a box to be "self-contained" if b contains its own label. Let C' be the set of all boxes that are *not* self-contained. Finally, let D be the set of all labels for boxes in C'. Then it immediately follows that D is not the contents of any single box. From this, it follows that there exists a set of naturals that is not equal to the contents of any box. So there are "more" sets of naturals than there are boxes. This is true *regardless* of how many boxes there are, or what naturals are contained by which boxes, as long as each box has a label. So we conclude: there are more sets of naturals than there are naturals. >2/ all possible digit sequences are computable to all, as in an infinite amount >of, finite lengths Once again, a completely incoherent mess. What is true, as I have already said, is this: For all reals r, for all natural number n, there exists a computable real r' such that r and r' agree on the first n digits. But the following, similar statement is false: For all reals r, there exists a computable real r' such that for all natural numbers n, r and r' agree on the first n digits. >ANY form of EVASION and not ANSWERING the questions will be discarded. You are a very poor student. You ask incoherent questions, and then expect to get coherent answers. Sometimes the correct is: Your questions are incoherent. -- Daryl McCullough Ithaca, NY
From: Tim Little on 8 Jun 2010 23:46 On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: > You seem to be backpedaling like the others now, that a computable > list is impossible anyway so I can't use it in my argument. There is such a concept as a computable list. Unfortunately for your pronouncements, no computable list can contain all of the computable reals. Just as certainly, no computable list can contain all the reals. - Tim
From: Daryl McCullough on 8 Jun 2010 23:47 George Greene says... > >On Jun 8, 8:22=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> Here's what's funny about USENET. In a regular classroom, >> you have one teacher and many students. In a typical USENET >> discussion, there are many teachers and just one student. >> You'd think that such a low student/teacher ratio would make >> for quick progress, but that doesn't turn out to be the case. > >This is true, but it's not for the reason you think. >The reason for the lack of progress is THE STUPID HALF >of the alleged teachers. I don't think that's correct. There is no teacher that is capable of explaining anything to Herc. -- Daryl McCullough Ithaca, NY
From: Sylvia Else on 8 Jun 2010 23:49
On 9/06/2010 10:22 AM, Daryl McCullough wrote: > Here's what's funny about USENET. In a regular classroom, > you have one teacher and many students. In a typical USENET > discussion, there are many teachers and just one student. A student is considered capable of learning, and sometimes actually wants to learn. I question whether the alleged one student fits the criterion. > You'd think that such a low student/teacher ratio would make > for quick progress, but that doesn't turn out to be the case. > Herc's ignorance can defeat any number of teachers, no matter > how knowledgeable and patient. Sylvia. |