Re: (Not quite) Cantor's diagonal proof

From: |-|erc (spam_at_fodder.abc)
Date: 10/27/04 

Date: Wed, 27 Oct 2004 10:48:03 GMT

"Dave Seaman" <dseaman@no.such.host> wrote in
> On Wed, 27 Oct 2004 05:27:11 GMT, |-|erc wrote:
> > "Barb Knox" <see@sig.below> wrote in
>
> >> But with "rational", not every infinite sequence of decimal digits
> >> represents some rational (e.g., all the non-repeating sequences, which
> >> represent non-rational reals). So the diagonal sequence need not represent
> >> a rational, so the attempted proof fails. Happy now?
>
> > You and Dave COMPLETELY missed the point of the question.
>
> > Here's another question. Suppose I list all of the repeating and
> > terminating decimals (i.e. rationals). What in Cantor's proof prevents me
> > from finding some sort of order for them so that I can use the diagonal
> > method to create a repeating decimal not in the list, thereby "proving" that
> > the rationals are uncountable? I know that I can't actually do this (the
> > rationals are countable),
>
> You have completely missed the point of my answer. Here it is again,
> from the other post:
>
> >> 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 you think this doesn't work, then you need to explain exactly where
> you think the problem is. Just waving your hands and claiming no such x
> exists doesn't cut it. That paragraph explains precisely why x exists
> and is not in the range of f.
>
> So the answer to your question is: the part of Cantor's proof that shows
> x is not in ran(f) is the part that prevents x from being in Q if
> Q \subseteq ran(f). Just apply the definition of "subset". That is not
> a circular argument.

what *are* you talking about?

> > Here's another question. Suppose I list all of the repeating and
> > terminating decimals (i.e. rationals). What in Cantor's proof prevents me
> > from finding some sort of order for them so that I can use the diagonal
> > method to create a repeating decimal not in the list, thereby "proving" that
> > the rationals are uncountable? I know that I can't actually do this (the
> > rationals are countable),

hint : *why* cannot the complete rationals be rearranged to form a repeating diagonal.
its not hard, and it doesn't require cantors proof.

Herc

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