Re: limitation to the halting proof

From: |-|erc (gotcha_at_beauty.com)
Date: 05/31/04 

Date: Sun, 30 May 2004 23:07:13 GMT

"Barb Knox" <see@sig.below> wrote in message > >>
> >> So where's the diagonal number in the list? Be specific.
> >> (Bear in mind that the diagonal number was constructed
> >> *from* the numbers in the list, as opposed to any
> >> generating function coupling them [Cantor does not require
> >> a generating function in his proof, merely an assumed list
> >> of reals], and need not be computable in the UTM sense.)
> >>
> >
> >The diagonal number itself appears only in the infinite list itself, not as a
> >member.
>
> So you DO concede that the diagonal number does not appear as a member of the
> list? If so, that's progress.

only when you can tell me the number, not a self referencing specification.

>
> >Tell me what digit is not in the list and I'll show you a result that contains
> >the diagonal number perfectly up to and including that digit.
>
> Big deal -- it's only the ENTIRE infinitely-long diagonal number that does not
> appear as a member of the list; it matters not a bit that some or all of its
> finite prefixes do appear as members.

lets examine the connundrum you are forced to endure. all the prefixes of the number
are on the list, of unlimited length, yet the entire string does not appear. how do
you sleep at night?

surely Cantor meant... it appears that a sequence of digits is not present. if you
have proven there is a missing sequence of digits then you can specify some finite
subsequence of the missing sequence that is also not present. Can you do that?

>
> >Stalemate.
>
> No (changing sports): own goal.

every list I declare you find a new number, every number you declare
I'll find it on that same list. you've just wrongly assumed your own benefit
of the doubt.

>
> >By induction the number is contained in the list for infinite digits.
>
> So you say, according to the following flawed reasoning:
>
> >We've been through this 100 times.
>
> And going through it a few more times won't make it any more valid.
>
> >0.1 is in the list. 0.12 is in the list. 0.123 is
> >in the list where diag is 0.123xyz.... for all digits in diag,
> >that number is in the list.
> >As the list approaches infinity the point on the number line is enscribed,
>
> Induction tells us that for every NATURAL NUMBER k, the k-digit prefix can
> appear as a member of the list. You then go on to conclude that for "K =
> infinity" this proves that the entire INFINITE sequence of digits appears as a
> member of the list.

not as a member.

This sequence of points in its infinite form will mark the point 1/3 [T / F]?

0.3
0.33
0.333
0.3333
...

>
> Is it possible that you might somehow someday understand that "infinity" is
> NOT a natural number? If you say it is, then maybe you can tell us which
> natural number is the one immediately before "infinity" (since we know that
> every non-zero natural number has a unique prececessor).

is it possible you might somehow understand that numbers are symbolic
represenatations of points and their specification of those points is merely that,
the string of digits is a symbol for the number. to refine the region on the
number line there are initially 10 options. that means whatever digit is used
there are 9 other candidates. This is all part and parcel of specifiying a number
using digit strings. Given the fact that there are 9 contender digits for anti-diag,
it doesn't matter, real numbers are a set and there are infinitely many that
start with the other 9 contender digits. Diag loses at position one, its not
possible to specify a 1 digit number that is not listable. Digit 2 is no different,
there are well over 100 reals we can list, so 0.00 to 0.99 are covered, Diag
loses again. Sets are specified on the foundation that their distinguishing members
are disjoint. All the digits are effectively identical in other regards. If you specify
a number 0.7xxxxxx you are *also* declaring 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.8, 0.9
as the vector space surrounding that path. It doesn't matter if diag is defined to
always select from that vector space, that's what the other members of the
SET are designed for. There is no digit in diag that forms a contradiction, even if
I present you with some finite list, the only contradiction you could try to predict
is off the list. If I give you a 1000000 by 1000000 list of digits, you can only suggest
the contradiction lies at digit 1000001 or higher. Numbers are options not sentences.

>
> But of course you can't answer that, because there is no such natural number,
> because "infinity" is not a natural number. Your induction is perfectly
> valid, but it only applies to natural numbers (which are each finite).

This is *your* induction.

x
this list is finite

x
xx
this list is finite

x
xx
xxx
this list is finite

therefore,

x
xx
xxx
...

this list is finite

>
> Now, you might want to read up on "transfinite induction", so you can spout
> mathematically-ignorant delusional rubbish in a whole new subject area...

all digit prefixes are on the list, therefore all_digits_are_on_the_list,
but the number is not on the list, now thats delusional rubbish.

Herc

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