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From: |-|ercules on 9 Jun 2010 01:52
(from the "Xenides dies" thread)

> For *all* N, the sequence differs from the Nth entry in the list at
> the Nth digit (and possibly other positions as well). It is new
> because for *every* sequence in the list, the question "is it the same
> as this sequence" is answered "no".
>

So you think the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS
and this does not contradict that ALL sequences of digits are on the computable
list of reals up to all (an infinite amount of) digit positions?

Herc
--
the nonexistence of a box that contains the numbers of all the boxes
that don't contain their own box number implies higher infinities.
- Cantor's Proof (the holy grail of paradise in mathematics)
From: William Hughes on 9 Jun 2010 02:36
On Jun 9, 2:52 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> (from the "Xenides dies" thread)
>
> > For *all* N, the sequence differs from the Nth entry in the list at
> > the Nth digit (and possibly other positions as well).  It is new
> > because for *every* sequence in the list, the question "is it the same
> > as this sequence" is answered "no".
>
> So you think the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS
> and this does not contradict that ALL sequences of digits are on the computable
> list of reals up to all (an infinite amount of) digit positions?
>

Yes saying "ALL sequences of digits are on the computable
list of reals up to all (an infinite amount of) digit positions"
does not mean that the list contains a sequence of digits that
does not have a last digit.


- William Hughes
From: Daryl McCullough on 9 Jun 2010 06:57
|-|ercules says...
>
>(from the "Xenides dies" thread)
>
>> For *all* N, the sequence differs from the Nth entry in the list at
>> the Nth digit (and possibly other positions as well). It is new
>> because for *every* sequence in the list, the question "is it the same
>> as this sequence" is answered "no".
>>
>
>So you think the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS
>and this does not contradict that ALL sequences of digits are on the computable
>list of reals up to all (an infinite amount of) digit positions?

You start with a completely crystal clear statement:
The antidiagonal number is not equal to any number on the list.

Then you paraphrase this clear statement to get a completely
muddled statement:

>the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS
>and this does not contradict that ALL sequences of digits are on
>the computable list of reals up to all (an infinite amount of)
>digit positions

Why do you prefer to use muddled, incoherent statements instead of
clear ones?

The antidiagonal is not equal to any of the numbers on the list.
What is unclear about that?

--
Daryl McCullough
Ithaca, NY

From: WM on 9 Jun 2010 07:11
On 9 Jun., 12:57, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote:

> The antidiagonal is not equal to any of the numbers on the list.
> What is unclear about that?

It is unclear how this can be accomplished in the binary tree with
countably many different paths which contains all numbers by
definition.

Regards, WM
From: |-|ercules on 9 Jun 2010 07:17
"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote...
> |-|ercules says...
>>
>>(from the "Xenides dies" thread)
>>
>>> For *all* N, the sequence differs from the Nth entry in the list at
>>> the Nth digit (and possibly other positions as well). It is new
>>> because for *every* sequence in the list, the question "is it the same
>>> as this sequence" is answered "no".
>>>
>>
>>So you think the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS
>>and this does not contradict that ALL sequences of digits are on the computable
>>list of reals up to all (an infinite amount of) digit positions?
>
> You start with a completely crystal clear statement:
> The antidiagonal number is not equal to any number on the list.
>
> Then you paraphrase this clear statement to get a completely
> muddled statement:
>
>>the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS
>>and this does not contradict that ALL sequences of digits are on
>>the computable list of reals up to all (an infinite amount of)
>>digit positions
>
> Why do you prefer to use muddled, incoherent statements instead of
> clear ones?
>
> The antidiagonal is not equal to any of the numbers on the list.
> What is unclear about that?
>

It's based on this argument.

123
456
789

DIAG = 159
ANTIDIAG = 260

260 is not on the list, it's a NEW DIGIT SEQUENCE.

You claim this works on infinite lists.

You claim no list contains EVERY DIGIT SEQUENCE
because you can find a NEW DIGIT SEQUENCE

But the computable real list contains EVERY DIGIT SEQUENCE
up to all (an infinite amount of) finite lengths.

EVERY DIGIT SEQUENCE POSSIBLE UP TO INFINITY!



BELOW IS A *VALID* DIAGONAL ARGUMENT

123
456
789

DIAG = 159
ANTIDIAG = 260


See how it actually generates a NEW SEQUENCE OF DIGITS!!!!!!!!!!!!!!!!

Your argument doesn't do that!

Here is what is ACTUALLY happening.

1 Start with a list containing all sequences.
2 Find a NEW sequence
3 CONTRADICTION

Herc

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