Why can no one in sci.math understand my simple point?
in [Logic]
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From: Virgil on 18 Jun 2010 15:32 In article <4c1b2c8e$0$1025$afc38c87(a)news.optusnet.com.au>, "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim(a)little-possums.net> wrote in message > news:slrni1m63n.jrj.tim(a)soprano.little-possums.net... > > On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > >> I made no premise. > > > > Sure you did: you assumed that no list of computable numbers can > > exist. You also assumed an incorrect definition of "computable". > > > > No, I assumed that a list of all computable numbers can exist. Then I gave a > simple algorithm which forms a computable number which is not on the list. I > therefore proved that no list of all computable numbers can exist. > > It is *exactly* the same as Cantor's proof that the Reals cannot be listed. > > It is of interest because it is known that the computable numbers are > countable. Therefore the property "cannot be listed" is *not* the same as > the property "is uncountable". > > Cantor's diagonal proof does *not* show the Reals are uncountable; it just > proves the much weaker statement that "the Reals cannot be listed". Given the axiom of choice, as in ZFC, any countable set must be, at least theoretically, listable, though such a listing need not be computable. And if countable, for infinite sets, does not mean bijectable with N (or listable), what does it mean?
From: Virgil on 18 Jun 2010 15:40 In article <88476fa9-bbc3-4ac1-9080-557f6a3b5f74(a)j8g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 18 Jun., 13:09, "Peter Webb" > > > The computable numbers are countable. > > > > And similarly Cantor's proof does not show that there are an uncountable > > number of Reals. > > What do you understand by "uncountable"? > > > It proves exactly what Cantor claimed it did, which is that > > you cannot list all Reals. > > Cantor said that there are 2^aleph_0 reals and aleph_0 rationals. And > he "proved" that 2^aleph_0 > aleph_0. And he said that there are an > uncountable number of reals because countable means listable. > > Regards, WM WM is right on this point!
From: Virgil on 18 Jun 2010 15:43 In article <4c1b5407$0$17174$afc38c87(a)news.optusnet.com.au>, "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim(a)little-possums.net> wrote in message > news:slrni1mcki.jrj.tim(a)soprano.little-possums.net... > > On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > >> Of course this number is computable; there > >> is a simple algorithm to compute it. > > > > I see you still haven't consulted a definition of "computable number". > > Umm, yes I have. > > > No worries, let me know when you have. > > > > > > - Tim > > There can be no list of all computable numbers. > > The proof is quite simple. Lets imagine that you have such a list of all > computable numbers. > > Lets say it starts off ... > > .111111... > .141592 ... > .71828 ... > > Take the 1st digit of the first number. If it is a "1", then make the first > digit of the diagonal number "2", otherwise make it a "1". Well, the first > digit of the first number is a "1", so the first digit of the diagonal > number is a "2". > > Now take the second digit of the second number and do the same substitution. > Its a "4", so the second digit of the diagonal number is "1". > > Now take the 3rd digit of the 3rd number ... its a "8", so the third digit > of the diagonal number is a "1". > > Continue in this fashion. > > The number that is produced is clearly "computable", because we have > computed it. Its also clearly not on the list. Therefore the list cannot > have contained all computable numbers. > > Exactly the same as Cantor's proof that the Reals cannot be listed. > > However, this does *not* mean that there are an uncountable number of them. > The computable numbers are countable. > > And similarly Cantor's proof does not show that there are an uncountable > number of Reals. It proves exactly what Cantor claimed it did, which is that > you cannot list all Reals. At least one definition of "countability" for infinite sets is "listability", i.e., existence of a surjection from N to the set in question. By what definition of countability is an infinite set which cannot be listed still regarded as countable?
From: Virgil on 18 Jun 2010 15:45 In article <b8f9104b-9daa-4892-9363-6ee43739c36a(a)a30g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > No. An infinite sequence of digits does not represent a number. In > general it does not even converge. Given a finite set of n digits and interpreting the sequence as an n-ary proper fraction, it ALWAYS converges. So the USUAL interpretation of such an infinite sequence DOES converge ALWAYS.
From: Virgil on 18 Jun 2010 15:46
In article <87r5k41dwf.fsf(a)phiwumbda.org>, "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote: > WM <mueckenh(a)rz.fh-augsburg.de> writes: > > > On 18 Jun., 14:17, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> Newberry <newberr...(a)gmail.com> writes: > >> > On Jun 15, 9:46�am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >> > wrote: > >> >> WM says... > >> > >> >> >On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> >> >> The proof does not make use of any property of infinite lists. > >> >> >> The proof establishes: (If r_n is the list of reals, and > >> >> >> d is the antidiagonal) > >> > >> >> >> forall n, d is not equal to r_n > >> > >> >> >As every n is finite, it belongs to a finite initial segment of the > >> >> >infinite list. > >> > >> >> I'm not sure what you are saying. The fact is, we can prove > >> >> that for every real r_n on the list, d is not equal to r_n. > >> >> That means that d is not on the list. > >> > >> > How do you know that it does not prove that an anti-diagonal does > >> > exist i.e. that an antidiagonal is a contradiction in terms? > >> > >> Because every infinite sequence of digits represents a real number? �And > >> the antidiagonal is one such sequence? > > > > No. An infinite sequence of digits does not represent a number. In > > general it does not even converge. In order to have convergence, you > > need the powers of 10 or 2 or so. But without a finite definition > > there are no infinite sequences at all, neither with nor without > > powers. > > Yes, very enlightening. What a unique and remarkable grasp of > mathematics. > > And what a shame that anyone allows you to teach. I concur! |