Re: limitation to induction on finite bounds

From: Barb Knox (see_at_sig.below)
Date: 06/24/04 

Date: Thu, 24 Jun 2004 16:45:41 +1200

In article <IToCc.59151$sj4.56471@news-server.bigpond.net.au>,
 "|-|erc" <gotcha@beauty.com> wrote:
[snip]

>Is this valid?

NO.

[snip]

>0
>0.3
>0.33
>0.333
>..
>
>This sequence contains the diagonal, 0.33..

As I posted out before, which you chose not to respond to:

Ah, I think I see the rule you're wanting to introduce: If one allows an
infinite list of sequences, then one MUST allow the individual sequences in
the list to be infinite themselves.

Trying to talk mathematics with you can be an intellectually interesting
exercise, rather akin to a chess player being introduced to fairy chess. So,
given your above non-standard "fairy maths" rule about infinite lists of
sequences, here's another go at showing that the infinite sequence .333... is
legitimately (by your rules) not in the infinite list .3, .33, .333, ...

First, let every member of the infinite list be an *infinite* sequence:
    . 3 0 0 0 0 0 ...
    . 3 3 0 0 0 0 ...
    . 3 3 3 0 0 0 ...
    . 3 3 3 3 0 0 ...
      ...

That is, the i'th sequence in the list consists of i 3s followed by an
infinite tail of 0s. Clearly, for every i, the i-digit prefix of 333...
occurs in the list (as the i'th member in fact). However, EVERY sequence in
the list has an infinite tail of 0s, which 333... conspicuously lacks.
Therefore 333... can not be in the list.

So, do you have a further fairy maths rule to exclude this construction? Or
do you now see that 333... is not in the list? Or maybe you'll just change
the subject. 

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