Why can no one in sci.math understand my simple point?
in [Logic]
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From: Mike Terry on 21 Jun 2010 16:30 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4c1eb078$0$1028$afc38c87(a)news.optusnet.com.au... > > "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote in message > news:dLGdnUdTkbKpzoPRnZ2dnUVZ8hKdnZ2d(a)brightview.co.uk... > > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message > > news:4c1e3743$0$1028$afc38c87(a)news.optusnet.com.au... > >> >> Perhaps if you could explain how my proof differs from Cantor's > >> >> proof, that would be a start. > >> > > >> > I have done already, but perhaps I can explain it in different terms > >> > for you. > >> > > >> > > >> >> You seem to accept Cantor's proof, but not mine. Yet they are almost > >> >> identical, the only difference being I have inserted the word > >> >> "computable" in front of "reals". > >> >> > >> >> I can't see how you can accept one and not the other. > >> > > >> > Cantor's proof shows that for any mapping L:N->R, antidiag(L) is real > >> > and not in range(L). Cantor's proof does include a proof that > >> > antidiag(L) is real. It does not show that antidiag(L) is a > >> > computable real. > >> > > >> > >> Yes. > >> > >> > You cannot just drop "computable" in there and suppose that the logic > >> > works, just as you cannot drop "rational" in there and suppose that > >> > the logic works. > >> > > >> > >> Well, it does work. > >> > >> > If you want to prove that for any mapping L:N->R, antidiag(L) is > >> > computable, you need to include a proof that antidiag(L) satisfies the > >> > mathematical definition for a computable real. > >> > > >> > >> OK. > >> > >> I an specify it to n places for all n using the following simple > > algorithm. > >> > >> Change the nth digit of the nth term from a 1 to a 2 and else to a 1. > >> > >> Thus if your list contained say: > >> > >> .1111111 > >> .2222222 > >> .1111222 > >> > >> The antidiagonal is 0.212... > >> > >> Notice that I computed this quite easily? > >> > >> Every item on the list is computable. So every item is known to at least > >> n > >> decimal places of accuracy, and Cantor gives us a very easy way to > >> compute > >> the first n terms on the anti-diagonal given the first n items on the > >> list > >> to n decimal places. > >> > >> > >> > > >> >> It is extremely easy to compute. > >> >> > >> >> Take the nth decimal place of the nth term. > >> > > >> > The nth decimal place of the nth term of what? The list L? That's > >> > fine if you permit infinite lists of reals as input to the algorithm, > >> > but the definition of "computable real" permits no such thing. > >> > > >> > >> No, to calculate the first n terms of the anti-diagonal you require only > > te > >> first n items on the list to n places of accuracy. > > > > So you require as input: > > - the 1st digit to 1 place of accuracy > > - the 2nd digit to 2 places of accuracy > > - the 3rd digit to 3 places of accuracy > > - ... > > > > The ... above is the give away! For your algorithm to work its going to > > require an INFINITE amount of input data. (Remember, you are trying to > > find > > a SINGLE stand-alone FINITE algorithm that produces ALL the required > > digits > > of the antidiagonal.) > > > > No, I only need a finite algorithm which will calculate it to n decimal > places with finite input. The definition of computable only requires me to > produce it to arbitrary accuracy. But obviously this single finite algorithm needs to be able to do this FOR ANY n. (I made an earlier post with more details.) > > I'm pretty sure you can't produce an algorithm which will produce *every* > digit in sqrt(2) in finite time. Sqrt(2) is computable because we can > calculate it to arbitrary accuracy, not because we can explicitly list every > digit in its decimal expansion in finite time; we can't. > > > > > So Tim was quite right, and you haven't answered his point correctly > > yet... > > > > Regards, > > Mike. > > > > >
From: WM on 21 Jun 2010 16:35 On 21 Jun., 22:17, Virgil <Vir...(a)home.esc> wrote: > > > But the set of all diagonals constructed according to the scheme given > > above is countable but cannot be listed. > > Whatever do yu mean by "the set of all diagonals"? > If you mean one "diagonal" for each possible list, that set of diagonals > is not countable. > If you mean the set of all "diagonals" for a single list, that is not > countable either, since each of uncountably many permutations of the > list produces a different "diagonal". Take a list of all rationals. Construct, accordingto a given substitution rule the antidiagonal. Add it at first position to the list. Construct, according to the given rule the antidiagonal, add it ... and so on. The number of antidiagonal is countable. Nevertheless it cannot be listed. > > > > > > Cantor has shown that the set of all reals cannot biject with the > > > naturals, ergo, the set of all reals is NOT of countable cardinality. > > > The same proof shows that some countable sets are of not countable > > cardinality. > > I have yet to see such a "proof" That is impossible, because you will not recognize it. But for the lurkers: The complete infinite binary tree contains, by definition, all real numbers between 0 and 1 as infinite paths, i. e., as infinite sequences { 0, 1 }^N of bits. 0, / \ 0 1 / \ / \ 0 1 0 1 / 0 ... The set { K_k | k in N } of nodes K_k of the tree is countable. K_0 / \ K_1 K_2 / \ / \ K_3 K_4 K_5 K_6 / K_7 ... The tree is constructed by extending the configurations B_j as explained below: _________________ B_0 = K_0 _________________ B_1 = K_0 / K_1 _________________ B_2 = K_0 / \ K_1 K_2 _________________ B_3 = K_0 / \ K_1 K_2 / K_3 _________________ B_4 = K_0 / \ K_1 K_2 / \ K_3 K_4 _________________ .... _________________ B_j = K_0 / \ K_1 K_2 / \ K_3 K_4 ... ... ... K_j _________________ .... _________________ The complete infinite binary tree (including all those infinite paths which consist of nodes and edges only) is constructed by a countable number of steps. In no step more than one infinite paths is extended. Hence there are not more than countably many infinite paths. Nevertheless these paths cannot be enumerated. Regards, WM
From: Virgil on 21 Jun 2010 16:38 In article <3d9bddf3-bc72-4377-a462-7afe5f0a700f(a)t10g2000yqg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 21 Jun., 12:41, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > Peter Webb says... > > > > >Cantor constructs the antidiagonal by taking the list and then computing a > > >missing Real based upon the decimal expansion of every Real on the list. > > > > That's right, so Cantor showed how to compute an antidiagonal > > using the list of reals as an input. That shows that the antidiagonal > > is computable *relative* to the list. It does not show that the > > antidiagonal is computable. "Computable relative to a list" and > > "computable" are two different things. > > > > This is a pretty simple concept, Peter. Mathematically, > > you have a function antidiagonal(L,n) which returns the > > nth digit of the antidiagonal, given the list L and n. > > That function is computable. But for the antidiagonal > > to be computable (not computable *relative* to L), you > > would have to be able to come up with a function > > > > antidiagonal(n) > > > > that, for *any* n, gives the nth digit of the antidiagonal. > > > > Two different concepts: computable relative to L, and computable. > > But none of them does help to save set theory. WM is the only who seems to have thrown away set theory, but, despite WM's urgings, almost no one else does. > > The diagonal numbers computable relative to a list are countable. And > so are the diagonal numbers computable relative to a list that is > extended in the n-th step by the diagonal number computed in step n - > 1. > > Nevertheless, there is not list of these "relative to the list" > computable diagonal numbers. Therefore Cantor's proof shows the > uncountability of a countable set. By at least some definitions of "computable number", the anti-diagonal of a list of computable numbers need not be itself computable, and if not, the set of computable numbers should be countable. And I am not at all sure that for definitions of "computable number" which would require such an anti-diagonal to be computable that the set of computable numbers is provably countable.
From: Virgil on 21 Jun 2010 16:53 In article <0e1c7c2c-32e9-4f67-b79f-502b81c21aa4(a)y11g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 21 Jun., 15:05, Sylvia Else <syl...(a)not.here.invalid> wrote: > > On 21/06/2010 7:51 PM, WM wrote: > > > > > > > > > > > > > On 20 Jun., 22:18, 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". > > > > > But this proof can be applied to this countable set and shows its > > > uncountability. > > > > Let's see. How might that work... > > > > OK, by way of example, take the set of rationals. It's countable, so we > > can list them. > > > > Now construct an anti-diagonal. It's clearly not in the list, so... > > > > ... um, well what, exactly? > > It is an irrational number. Add it at first position to the former > list L0, obtain L1. Now construct the diagonal of L1 (according to a > fixed scheme). It is certainly an irrational number. Add ist to the > list L1 at first position, obtain list L2. Now construct the > diagonal, ... and continue. I this way you get a countable set > consisting of all rationals and of all diagonal numbers of these > lists. This set is certainly not countable, because you can prove that > there is always a diagonal number not in the list. On the other hand, > the set is countable by construction. What now? In what position in each new list is the anti-diagonal of the immediately previous list to be inserted? While any finite number of such insertions will indeed create new lists, will an infinite number of insertions work? For example, if the new entry is always placed first in the new list, the after infinitely such insertions there will be original members with infinitely many predecessors, so that one no longer has a list. And one clearly cannot put any new element after all others, as that also destroys the "being a list" property. Thus WM's arguments again fall flat and again demonstrate his inability to think things through.
From: Transfer Principle on 21 Jun 2010 17:50
On Jun 21, 5:46 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Newberry <newberr...(a)gmail.com> writes: > > What I had in mind is that if you drop the axiom of extent you can > > either conclude that there is no such list or that the anti-diagonal > > does not exist analogously to the conclusion that the set R = {x | ~(x > > in x} does not exist. > I'm not sure what the axiom of extent is. It should be evident to Hughes that "Axiom of Extent" refers to the "Axiom of Extensionality," as "extensionality" is related to the English word "extent." A quick Google search -- which Hughes could have done himself in seconds -- reveals sources using the name "Axiom of Extent" that even Hughes should consider respectable, including papers written by mathematicians, textbooks (via Google books), and the class websites of some university math classes. Newberry discusses dropping Extent (Extensionality). He might be interested in a current zuhair thread, where zuhair also considers the consequences of dropping this axiom. |