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From: |-|ercules on 17 Jun 2010 21:53
CONSTRUCT a *new sequence of digits* (somehow?) like so

An AD(n) =/= L(n,n)

Then we PROVE that it's a new sequence of digits

An AD(n) =/= L(n,n)

VOILA! Superinfinity! Just ignore the fact that it never comes up with a new sequence of digits.



Consider this LOGIC LOOP as an infinite sequence of propositions.

AD = 0
D = 1
WHILE TRUE
AD = AD + a different digit than the Dth digit of the Dth real) / 10^D
YOU HAVEN'T PRODUCED A NEW DIGIT SEQUENCE YET!
D++
WEND

How does this make a *new digit sequence* that is not present on the
list of computable reals?



Herc
--
If you ever rob someone, even to get your own stuff back, don't use the phrase
"Nobody leave the room!" ~ OJ Simpson
From: |-|ercules on 19 Jun 2010 05:34
CANTOR'S PROOF of modifying the diagonal does not create any new sequence at all.

All it does is this

CONSTRUCT a digit sequence like so
An AD(n) =/= L(n,n)

And then you say, it's different to each number like so

PROOF
An AD(n) =/= L(n,n)

But you have not demonstrated a NEW SEQUENCE OF DIGITS.

All you've done is this

[ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] -> Higher Infinities

Your actual 'proof' is a specific example of the above 'proof'!

[ An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n) ] -> Higher Infinities

Does anyone agree with the above version of Cantor's proof?

Herc
From: George Greene on 19 Jun 2010 14:54
On Jun 17, 9:53 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> CONSTRUCT a *new sequence of digits* (somehow?) like so
>
> An AD(n) =/= L(n,n)

That is NOT a construction. That is just an ASSERTION ABOUT
each individual element of the anti-diagonal. And it is mis-stated in
any case.
AD DOES NOT TAKE (n) as an argument! AD(.) is the anti-diagonal OF A
LIST!
It takes L as an argument. To CONSTRUCT this, you would have to say
what the nth
element of the anti-diagonal WAS EQUAL to,NOT what it was UNequal to!
What you ACTUALLY MEANT to say here was
An[ AD(L,n) = 9 - L(n,n) ]
THAT is a construction.
The statement you made will follow FROM this as a consequence, since
Ane{0..9}[ 9 - n =/= n ]


> Then we PROVE that it's a new sequence of digits
>
> An AD(n) =/= L(n,n)

Well, yes, this does prove that the anti-diagonal of L is NOT ON the
list L.
But that actually does NOT mean that ANYthing is NEW! It just means
that
the list DID NOT ALREADY HAVE this element on it! It means that L WAS
NOT
"the list" (or even A list) of EVERY real!
However, since we didn't make any assumptions at all about L here,
other than
that it was square and denumerable (in both directions), this proves
THAT ANY AND EVERY
list WITH THAT SHAPE is NOT "the list of all" reals!
In other words, THERE CANNOT BE ANY "list of ALL" reals!


> VOILA! Superinfinity!

This is not where the superinfinity part comes from.
It is possible that the class of all reals either a) doesn't exist, or
b) even if it does, does not HAVE
a SIZE. Actually getting to a superinfinity number REQUIRES MORE than
this, but since
you already had a list that was both infinitely wide and infinitely
long, the axiom of infinity
has sort of been assumed already. You are going to need to invoke
choice and power as well,
to get an actual new number, but this is of course all above your
head.

> Just ignore the fact that it never comes up with a new sequence of digits.

AD(L) *IS* a new sequence, DUMBASS.
YOU JUST PROVED it was NEW!
It WAS NOT ON the ORIGINAL list L!

>
> Consider this LOGIC LOOP as an infinite sequence of propositions.
>
> AD = 0
> D = 1
> WHILE TRUE
>    AD = AD + a different digit than the Dth digit of the Dth real) / 10^D
>    YOU HAVEN'T PRODUCED A NEW DIGIT SEQUENCE YET!
>    D++
> WEND
>
> How does this make a *new digit sequence* that is not present on the
> list of computable reals?
>
> Herc
> --
> If you ever rob someone, even to get your own stuff back, don't use the phrase
> "Nobody leave the room!" ~ OJ Simpson

From: |-|ercules on 19 Jun 2010 16:21
"George Greene" <greeneg(a)email.unc.edu> wrote
> On Jun 19, 5:34 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
>> CANTOR'S PROOF of modifying the diagonal does not create any new sequence at all.
>>
>> All it does is this
>>
>> CONSTRUCT a digit sequence like so
>> An AD(n) =/= L(n,n)
>>
>> And then you say, it's different to each number like so
>>
>> PROOF
>> An AD(n) =/= L(n,n)
>
> THAT'S RIGHT, DUMBASS, IT IS DIFFERENT from EVERY number on the list!
>
>>
>> But you have not demonstrated a NEW SEQUENCE OF DIGITS.
>
> YES, WE HAVE! THIS NUMBER IS NOT ON the OLD list!
> THEREFORE, IT IS *NEW*!! YOU HAD SAID that the OLD list was
> the list of ALL the numbers! IT WASN'T! THIS NUMBER was NOT ON it!
> THIS NUMBER IS THEREFORE *NEW*!!


Rubbish! You don't show any new pattern of digits, you just SAY it's new.


If all digits of a single infinite expansion can be contained with increasing finite prefixes,
and the computable set of reals has EVERY finite prefix, then all digits of EVERY infinite
expansion are contained (in increasing segments).

Find some suitable substitution for the word "contained" since this bothers you, and
that proposition disproves that modifying the diagonal produces a new sequence of digits.



Yes I know

An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n)

and millions of other zero entropy formula.

Herc
From: George Greene on 20 Jun 2010 02:36
On Jun 19, 4:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:

> Yes I know
>
> An AD(n) = (L(n,n) + 1) mod 9

No, you don't.
ACTUALLY,
AD(n) = 9 - L(n,n)

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