Re: (Not quite) Cantor's diagonal proof
From: Dave Seaman (dseaman_at_no.such.host)
Date: 10/27/04
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Date: Wed, 27 Oct 2004 06:17:27 +0000 (UTC)
On Wed, 27 Oct 2004 04:20:19 GMT, |-|erc wrote:
> "Dave Seaman" <dseaman@no.such.host> wrote in
>> On Tue, 26 Oct 2004 09:06:20 GMT, |-|erc wrote:
>> > "Dave Seaman" <dseaman@no.such.host> wrote in
>> > *a version of*
>> >> The diagonal argument depends on the least upper bound property of the
>> >> real numbers. The diagonal argument produces a decimal digit string,
>> >> which is associated with a certain infinite series. The partial sums of
>> >> that series form a set of real numbers that is nonempty and bounded
>> >> above. The least upper bound of that set is the required number.
>> >> The same argument does not work when applied to the rationals, since the
>> >> rationals do not satisfy the LUB property.
>> >> And, by the way, there is nothing circular about the argument that if the
>> >> original list contains all of the rational numbers, then the number
>> >> produced by the diagonal argument, which is necessarily different from
>> >> any number in the list, must therefore be irrational.
>> > Diagonalisation is a valid technique because diagonalisation is a valid technique.
>> > I'm getting used to this argument!
>> The diagonal argument works because, given a mapping f: N -> R, the
>> argument produces a number x whose n'th digit differs from the n'th digit
>> of f(n) for each n. And since x is chosen to avoid dual-representation
>> problems, it follows that x is not in the range of f.
> if such an x was possible, which it isn't.
Why not? Do you mean the diagonal fails to produce a decimal digit
string, or the decimal digit string produced by the diagonal process
fails to specify a real number?
Which is it? In either case, you're wrong, but I can't tell where your
error lies without further information.
>> > Such a proof gives no reassurance of the diagonalisation technique, which was the whole point
>> > of the OP. Assume diagonalisation may not work, then consider the example of rationals.
>> > 1 Start with a complete list of rationals
>> > 2 Order the list to have diagonal 0.12345678901234567890...
>> > Every digit is present on the diag so no matter what digits your rational has it can
>> > slide up or down the list until it fits.
>> Consider the number 0.098765432109876543210987654321.... It's a rational
>> number. Where does it go in your list?
> Why look for an error in this demonstration when the number
>> >> produced by the diagonal argument, which is necessarily different from
>> >> any number in the list, must therefore be irrational.
Would you have believed me if I had claimed that there must be an error
somewhere in your argument without actually pointing it out? That is
your tactic, not mine.
>> > 3 Make an anti-diag, this has 10(n) repeating digits so it is rational but is not on the list.
>> So, your method works because it works? Very clever.
> theorems relate to other theorems in a directed acyclic graph, not a heirarchy, try some different routes.
You are confused. The person who needs to try again is the person whose
argument has failed. That would be you, since your attempted list of
rationals is not complete.
-- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
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