Why can no one in sci.math understand my simple point?
in [Logic]
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From: Peter Webb on 20 Jun 2010 21:38 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1td72.jrj.tim(a)soprano.little-possums.net... > On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> "Tim Little" <tim(a)little-possums.net> wrote in message >>> However, there is no way that you can then prove the existence of a >>> finite algorithm accepting only the *finite* input n and producing the >>> n'th antidiagonal digit. >> >> Of course I can prove the existence of a finite algorithm. I can produce >> it. >> >> To produce the nth decimal of the anti-diagonal, look at the first n >> items on the purported list of all computable Reals. > > In what way is that "a finite algorithm accepting only the *finite* > input n"? Its finite, and its an algorithm. >It either must embed the list within the algorithm (thus > rendering it not finite), No, it embeds the first n digits of the first n terms to calculate the first n decimal places, and does for all n. That a finite process can be used to determine a number to any required degree of accuracy makes it computable. > or accept the list as additional ainput > (thus violating the condition that it accept only the finite input n). > > >> Well, I think I am making a pretty good hand of my case. But yes, I >> know what I am saying is not accepted. I am genuinely confused by >> this - the fact that my opinions are unique, and I would expect >> therefore to be almost certainly wrong (with probability 1). On the >> other hand my argument seems pretty solid to me. > > Read through very carefully the definition of a computable real, > perhaps also reading some of the other known results on computable > reals and following their proofs. At some point you will hopefully > see why the "construction" in Cantor's diagonal proof does not qualify > as a finite algorithm in the sense of the definition of computable > real. > Its an explicit construction. I can even do it by hand to tell you the anti-diagonal to any required degree of accuracy in a finite number of steps. That makes it computable. > > - Tim
From: Virgil on 20 Jun 2010 21:43 In article <b1d23660-1b7a-4682-affa-952cf2d9e03b(a)e34g2000pra.googlegroups.com>, Newberry <newberryxy(a)gmail.com> wrote: > On Jun 20, 1:18�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <f92c169d-ee85-40c2-aa82-c8bdf06f7...(a)j4g2000yqh.googlegroups.com>, > > > > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 20 Jun., 17:51, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > > > > > Cantor was this utterly insane freak who chose not to accept Newberry's > > > > word for it, and instead *prove* that there was no list of all real > > > > numbers. Obviously, his proof is nonsense, because, after all, Newberry > > > > said there was no list. > > > > > His proof is nonsense because it proves that a countable set, namely > > > the set of all reals of a Cantor-list and all diagonal numbers to be > > > constructed from it by a given rule an to be added to this list, > > > cannot be listed, hence, that this indisputably countable set is > > > uncountable. > > > > That is a deliberate misrepresentation of the so called "diagonal proof". > > > > What that proof (not actually by by Cantor himself) ays is that any > > listing of reals is necessarily incomplete because given such a listing > > one can always construct a real not listed in it. > > Please CONSTRUCT the anti-diagonal in the space provided below. > > Virgil's anti-diagonal construction > BEGIN It is not mine but Cantor's, though the following notation is mine. N represents the set of natural numbers. NxN is the Cartesian product of N with itself with members (m,n) where m and n are members of N. F: NxN --> {0,1} is a list of functions, F(m,--), from N to {0,1} with, for each m in N, F(m,--): n --> F(m,n), being such a function. Then the anti-diagonal function g:N --> {0,1} is defined by g(n) = 1-F(n,n) in {0,1}, so that g(n) <> F(n,n) for all n, and thus g(--) <> F(n,--) for all n.. Note the Cantor himself never published an anti-diagonal argument for the reals themselves, though some of his followers did. > > > > END > > > What Cantor himself proved was that for any list of infinite binary > > sequences there are binary sequences not listed in that list. > > > > Cantor had previously proved by quite a different method, which WM seems > > unable or unwilling to criticize, that the set of reals was not > > countable. > > > > > > > > > Even completely blinded matheologicians should be able, perhaps after > > > some contemplation, to recognize that. > > > > What completely blinded matheologicians are able, perhaps after > > some contemplation,to recognize is that WM is not working in the same > > world as mathematicians do.
From: Transfer Principle on 20 Jun 2010 21:45 On Jun 20, 8:51 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Newberry <newberr...(a)gmail.com> writes: > > That's right. There is no formula or algorithm to construct the list. > > It means that you would have to flip each and every digit one by one. > > And that is impossible. > Of course, it is perfectly reasonable to accept that there is no such > list, because Newberry says so. Ah, so it's nice to see that Newberry has joined Herc and WM in this ever-growing thread. > Cantor was this utterly insane freak who chose not to accept Newberry's > word for it, and instead *prove* that there was no list of all real > numbers. But this proof that there is no list of all real numbers must use the axioms of some _theory_ (such as ZFC) -- a theory which Newberry isn't required to accept just because Hughes says so. So Hughes continues to be a Herc-religionist (i.e., a religionist according to Herc), appealing to ZFC to prove Newberry wrong even though there's no reason to assume that Newberry is working in ZFC (or any other theory proving Cantor's Theorem).
From: Transfer Principle on 20 Jun 2010 21:49 On Jun 20, 12:09 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > "Sylvia Else" <syl...(a)not.here.invalid> wrote > > I've told you which step I have reservations about. > What? What a [expletive] womans answer that is. And so Herc plays the gender card here. I don't like it when anyone appeals to sexism to make an argument, regardless of which side of the debate he is on. It's as wrong for Cooper to make fun of Else because she's female as it is for others to make fun of galathaea because she's female. I applaud Herc for standing up to those whom he considers to be "religionists," but it hurts his cause when he makes sexist posts.
From: Peter Webb on 20 Jun 2010 21:50
"Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1teqk.jrj.tim(a)soprano.little-possums.net... > On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote in >> message >>> Not so - the "very simple algorithm" you are suggesting requires >>> "the input list" in its raw (infinite) form as input. >> >> No. Computing the the anti-diagonall to n places only requires the >> first n items on the list to n digits of accuracy. > > Read the definition of computable real carefully, please! > > You need to show that there exists a finite algorithm that accepts > input n, with *no* additional input, and produces the n'th digit. > Yes. Cantor produced such an algorithm. It is finite; the number can be specified to any arbitrary degree of accuracy in a finite number of steps with only a finite input. Hell, I can do it by hand, so it must be finite. Give me your list of all computable Reals and I will compute the anti-diagonal to arbitrary precision for you. > You don't get to put an upper bound on n when exhibiting the > algorithm. The same algorithm must work for all n. > The same algorithm does work for all n. Look at the first n terms to n places of accuracy. > For some fixed M, you could embed M digits of the diagonal into the > algorithm, and the algorithm would work for all n < M. But that's not > enough: it has to work for all n. > I can in fact compute the nth digit of the anti-diagonal for any n by hand. Cantor's original proof has the same requirements as mine. He asks for a purported list of all Reals and shows how to construct a missing Real. So do I. Of course I need to see the list first, and use this infirnite list in my proof, or else I couldn't tell you which Real is missing. But so does Cantor .... > >> How is this number any less computable than the square root of 2? > > There exists a straightforward algorithm (e.g. Newton's method) to > calculate square roots to arbitrary precision. You can embed the > value 2 into an algorithm which accepts n as input, and applies > Newton's method to generate the first n digits of sqrt(2). Given a purported list of all computable Reals, I can explicitly construct a Real not on the list to any required degree of accuracy. > > You cannot embed the list L into an algorithm which applies the > antidiagonal procedure for arbitrary input n, as the list L is > infinite. You can only embed finitely many digits of L, and so there > will be an upper bound on the values of n for which the algorithm > works. The same objection could be raised to Cantor's proof; after all, he accepts a purportedly infinite list as input to his algorithn as well. > > >> I don't use the whole list. To calculate the nth digit of the >> anti-diagonal I only need the first n terms to n decimal places >> accuracy. That is a finite amount of data. > > However, you then need infinitely many different algorithms for > increasing n. A number is only computable if the same algorithm works > for all n. > Well, the infinite list must be supplied in advance. Or else I can't work out what is missing. But then, Cantor's original proof has exactly the same rtequirement. > >>> Now IF you could codify all this data somehow into a finite >>> program, so you had a computable function D(m,n) which produced the >>> n'th digit of the m'th number in the list, you would be OK. But >>> you DON'T have such a computable function D. >> >> Well, Cantor asks for a list of Reals. And D(m,n) is computable in >> my case by definition > > No, it is not. The m'th entry is computable for each m, with a > different algorithm for every m. For D to be computable, there must > exist a *single* finite algorithm that subsumes them all. There > exists no such algorithm. > Well, the algorithm is clearly computable. If you give me a purported list of all computable numbers, I can compute a missing number. Or are you saying I have to do this without using the infinite list itself? I note that Cantor's original proof also requires an infinite list of Reals, but you seem to have no problem with that. > > - Tim |