Re: Cantor reloaded

From: The Ghost In The Machine (ewill_at_sirius.athghost7038suus.net)
Date: 12/07/04 

Date: Tue, 07 Dec 2004 05:00:08 GMT

In sci.math, Dave Seaman
<dseaman@no.such.host>
 wrote
on Mon, 6 Dec 2004 14:58:44 +0000 (UTC)
<cp1s34$p5n$1@mailhub227.itcs.purdue.edu>:
> On 6 Dec 2004 07:29:46 -0600, thdmuync wrote:
>>> albrechtwrote:
>
>>> So you would argue in the same way: The conclusion of Cantor's
>>> diagonal proof (on reals) is either: no (complete) list exists, or:
>>> the antidiagonal does not exists. ?
>
>>> Albrecht
>
>> No, actually by thinking about it, I think the conclusion is more
>> subtle (makes you think doesn't it :-) )
>
>> How do we define the antidiagonal. Well, the antidiagonal is that
>> number that differs at the i th diagonal position with the i th
>> diagonal position of the i the number in the list.
>
> There are many such numbers. The antidiagonal is one of them.

An uncountable number, in fact, as one can diagonalize
the antidiagonal list in the following fashion.

Let L be the (alleged) list of real numbers, implicitly setting
up a mapping from N to [0,1). Let l_n be the n'th element
on that list, and l_n(j) the j'th digit of l_n.

If one defines d_n(j):

d_n(j) = 7 if j == n and l_n(n) == 6
d_n(j) = 6 if j == n and l_n(n) != 6
d_n(j) = 5 if j != n and l_j(j) == 4
d_n(j) = 4 if j != n and l_j(j) != 4

it's fairly obvious that I now have a countably infinite
list D, and now I can diagonalize *that* as well, defining
a new number e. Done right, e is in *neither* list.

This sort of thing could go on indefinitely (L and D,
after all, have Lebesgue measure 0) although one might
have to take the digits two, three, ... at a time as one
continues the process, and of course one has to take some
care around 0, 9, 00, 99, 000, 999, ... .

Cantor's first proof is probably a more elegant method
to prove c != aleph_0, but this works. :-)

[rest snipped] 

-- 
#191, ewill3@earthlink.net
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