From: Graham Cooper on
> Consider the list of computable reals.

> Let w = the digit width of the largest set
> of complete permutations

> assume w is finite
> there are 10 computable copies of the
> complete permutations of width w
> each ending in each of digits 0..9 (at position w+1)
> which generates a set larger than width w
> so finite w cannot be the maximum size

> therefore w is infinite
> ----

you should recognize this form of induction

no maximum Natural number implies
there are infinite quantity of Natural numbers

IS TO

no maximum digit width of all full permutation sets implies
there is infinite digit width of all permutations
(of computable reals)

THEREFORE

modifying the diagonal of the list of computable reals
does not construct a new digit sequence

Herc
From: George Greene on
On Jun 24, 6:50 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote:
> > Consider the list of computable reals.

NO. You may NOT consider that.
In the first place, THAT list is itself NOT computable.
In the second place, EVERYthing you are WANTING to claim about
that list can be JUST as correctly claimed about the list of all
FINITE
bit-strings (which IS computable).

> > Let w = the digit width of the largest set of complete permutations

No. w may NOT equal that, since THERE IS NO SUCH THING as a
"permutation" for purposes of this question. What you are TRYING TO
CALL
a "permutation" IS JUST A SEQUENCE -- IS JUST A LIST, and every one of
them
IS DIFFERENT AND SEPARATE IN ITS OWN RIGHT. THERE IS NO important
sense in which some of them are permutations of others, and even if
there were,
YOU WOULDN'T KNOW WHICH to call the "base" or identity permutations
and
which to call "permuted". A sequence is a sequence. A list is a
list.
The fact that it is a permuation of its other permutations IS NOT
relevant!

More to the point, if you are talking about reals,
ABSOLUTELY ALL OF THEM, ALL THE TIME, are of width
w
IF they are going to be expressed as digit-strings.
So there is no point in talking about ANY OTHER width,
or asking about "the largest set": EVEN THE SMALLEST non-empty
set, the one with ONE real, IS OF WIDTH W.
No matter how many more reals you may add to get a bigger set,
SINCE THEY WILL ALL, ALWAYS, BE OF WIDTH w,
w
IS THE ONLY width that will ever be relevant to this question.
And w is A CONSTANT. It is NOT a variable or a statistic or something
that you might need to compute. It is KNOWN EXACTLY, IN ADVANCE.



From: Graham Cooper on
On Jun 25, 8:50 am, Graham Cooper <grahamcoop...(a)gmail.com> wrote:
> > Consider the list of computable reals.
> > 
> > Let w = the digit width of the largest set
> > of complete permutations
> > 
> > assume w is finite
> > there are 10 computable copies of the
> > complete permutations of width w
> > each ending in each of digits 0..9 (at position w+1)
> > which generates a set larger than width w
> > so finite w cannot be the maximum size
> > 
> > therefore w is infinite
> > ----
>
> you should recognize this form of induction
>
> no maximum Natural number implies
> there are infinite quantity of Natural numbers
>
> IS TO
>
> no maximum digit width of all full permutation sets implies
> there is infinite digit width of all permutations

I defeated myself here...

There are infinite amount of digit widths. One could argue an
infinite amount of finite digit widths!

But I think "infinite maximum digit width" supports
the claim of all possible sequences.

Herc
From: George Greene on
On Jun 24, 6:50 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote:
> > Consider the list of computable reals.
> > 
> > Let w = the digit width of the largest set
> > of complete permutations
> > 
> > assume w is finite

This is just idiotic.
Every real is infinitely wide BY DEFINITION,
and again, this width is, BY DEFINITION, THE SMALLEST infinity,
so w is KNOWN IN ADVANCE NOT to be finite.
From: Dingo on
On Thu, 24 Jun 2010 15:50:51 -0700 (PDT), Graham Cooper
<grahamcooper7(a)gmail.com> wrote:

>Herc

Oh please let it be.....and if you'd said your final thread altogether
that would be even more gooderer.....