Re: abundance of irrationals!)

From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 03/08/05 

Date: 8 Mar 2005 12:32:58 -0800

W. Mueckenheim wrote:
> But you said that the diagonal is different from any number in a
line.
> There we have all numbers with a finite sequence of digits 1. The
only
> possibility for the diagonal to distinguish itself from all such
> numbers is to have infinitely many digits 1. Is infinitely many not
> more than any finite number?

Yes.

Every number in the list has the same (countably infinite)
number of digits.

But they all have only finitely-many 1's.

>
> >
> > I'm saying that your attempt to construct the diagonal
> > as the limit of a sequence proves nothing,
>
> Why do you think the diagonal cannot be considered the limit of its
> segments?

I didn't say that:

(1) It doesn't need to be.
(2) By attempting to so construct it, you claim a problem.
But the "problem" is your misunderstanding of what a
limit is.

It's not that it can't be so constructed, it's just that
including limits, which you don't understand, into
Cantor's construction, which you don't understand, doubles
the number of things for you to get wrong.

> > Try to follow. I'll show you the parallel logic:
> >
> > Every element in the sequence {1/n, n=1,2,3,...}
> > is nonzero. The limit of the sequence is 0. 0 is not
> > in the list.
>
> And the corresponding limit is not on the diagonal.

Uh, yes. It's a number that's not in the list. Therefore
the list is incomplete.

> >
> > Every element in your sequence is different from 1/9.
> > The limit of the sequence is 1/9. 1/9 is not in your
> > list.
>
> And it is not on the diagonal.

It's a number that's not in the list. Therefore
the list is incomplete.

Actually, it *is* the diagonal. Consider: the
diagonal is a number which has a 1 in every position
n, for n=1,2,3,....

1/9 is a number which has a 1 in every position n,
for n=1,2,3....

Now you say 1/9 is not the diagonal. Please tell me
in which way the diagonal differs from 1/9.

             - Randy

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