Re: limitation to the halting proof

From: The Ghost In The Machine (ewill_at_aurigae.athghost7038suus.net)
Date: 05/29/04 

Date: Sat, 29 May 2004 20:00:11 GMT

In sci.logic, |-|erc
<gotcha@beauty.com>
 wrote
on Sat, 29 May 2004 10:02:04 GMT
<wcZtc.16537$L.3710@news-server.bigpond.net.au>:
> "The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net>
>> > can *write* hence *see*. you use a faulty premise of open
>> > diagonalisation and when its shown to be nonsense you shift
>> > the burden.
>>
>> I've yet to see a working proof that it's nonsense. Granted, every
>
> you have. a consistent enumeration of all functions is entirely possible.
> [T / F] now its not enough diag. is invalid, the proof that enumeration
> is impossible you all flaunt is gone, so you change your stance to
> "ok I want to see the actual list of numbers". it like proving the
> Earth is round, but no you want to see the rewards of trade before
> you send off a boat.

I still want to see the list of numbers. In the case of Q+, such
a list is possible:

1 <-> 1/1
2 <-> 1/2
3 <-> 2/1
4 <-> 1/3
2/2 (duplicate)
5 <-> 3/1
6 <-> 1/4
7 <-> 2/3
8 <-> 3/2
9 <-> 4/1
10 <-> 1/5
2/4 (duplicate)
3/3 (duplicate)
4/2 (duplicate)
11 <-> 5/1
12 <-> 1/6
13 <-> 2/5
14 <-> 3/4
15 <-> 4/3
16 <-> 5/2
17 <-> 6/1
...
1/n (n, k in N, n > 0, 0 < k <= n)
2/(n-1)
3/(n-2)
...
k/(n-k+1)
...
n/1
...

This list is of course not well-ordered in the usual sense
(for any rational number p/q, I can find p/2q closer to 0,
or (q+p)/2q closer to 1). Nevertheless, it establishes
a 1-1 surjective correspondence between N and Q, and with
a little work can be made bijective, by throwing out the duplicates.
One can also change Q+ into Q, by doing tricks such as the
following:

1 <-> 0
2 <-> 1/1
3 <-> -1/1
4 <-> 1/2
5 <-> -1/2
6 <-> 2/1
7 <-> -2/1
8 <-> 1/3
9 <-> -1/3
etc.

As a side issue, one is reminded of ferrule core memory, for those
of us old enough to remember such; the enumeration threads through
Q in a quasi-diagonal fashion.

>
>> > you have no mathematical opposition to countable functions and
>> > countable reals is just a step further. UTM(n), from n e N IS
>> > the countable reals, it doesn't miss a beat, not all n halt thats
>> > a petty copout requirement of YOUR argument, now even that is invalid.
>>
>> It's not even countable. It's finite.
>
> N is infinite, its the only infinity.

The set of all usable N is finite. Of course you're right in that
N is countably infinite; Cantor has proven (twice) that R is not.

> infinitely long lists of all variations of computable functions
> (in some consistent dogma if you insist it completely halts aswell)
> scribe the entire number line. the list is infinitely long its
> impossible to miss a permutation, NO MATTER HOW LONG IT IS.

Really?

That doesn't seem to square with Cantor's first proof.

http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof

The sequence x_i need not be computable or have finite digit
representations, but it's clear that it doesn't even exist anyway.

> even infinite permutations, even the diagonal modified infinite
> permutation appears
> within_the_infinite_list, enscribing that number onto the number line in
> perfect precision. its difficult to *index* that number because of all
> the cyclic tricks in its definition, but in the infinitely long list
> all digits of the modified diagonal appear. ALL

Yeah, sure it does. Cantor's first proof doesn't rely on digit expansions.

>
>
>> Of course mathematics does
>> not always deal in the concrete and the graspable --
>
>
> no there is no "of course" here, this is laymens language of
> yesteryears fairy numbers.

So R must be limited to only computable numbers? An interesting notion,
which basically renders R finite.

>
> Herc
> 

-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.

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