Re: Request for Peer Review - Refutation of Cantor Theorem Conclusion


Scott wrote:
George:

It has a *1* in position 2 because
what-you-now-have-as-item-2 was DEFINED to be
the anti-diagonal of the ORIGINAL set, and the ORIGINAL
set had, as ITS item 2,

The matrix is the original set, P(S), so #2 is in the set.

Bull***.
The original set/matrix IS NOT P(S).
It is your lame-ass ATTEMPT at P(S), and it FAILS to be P(S)
precisely BECAUSE it does NOT contain #2.
It CANNOT contain #2 because #2 is DEFINED
as DIFFERING from EVERY element of the original set.
It differs from the nth item-on-your-list in the nth bit-column,
FOR ALL n.

Simply removing it

Nobody is talking about removing anything.
You started with your whole list.
What we proved was that we could ADD something NEW to it.
means you are working with a subset of P(S) and not P(S)
Show me the algorithm on the set provided; ie, P(S).

You HAVE NOT provided P(S).
You HAVE provided A LIST of bit-strings.
A COUNTABLE list of bit-strings.
A list with the SAME length as S (or as N, since
S *is* N, for your purposes here, although the argument
holds for any set). No list of n items can contain 2^n
elements. 2^n IS BIGGER than n. You will ALWAYS
be able to diagonalize. N does not even need to be
infinite for this proof to work. If you have a set of
only FIVE elements (5 = {0,1,2,3,4}) then any list of
FIVE bit-strings is NOT going to be ALL the bit-strings
(because there are 32 of those). You give me any list
of 5 bit-strings-of-length-5, I can give you a new bit-
string not on your list. I can give you 27 of them in fact,
but that is not the point. The point IS that as long as you
only have 5, I can find a 6th, so there MUST be more than 5
of them.


No, it doesn't. The contradiction cannot occur at
any individual place. The contradiction occurs BECAUSE
there is diagonal disagreement at ALL INFINITELY many places
(along the diagonal) SIMULTANEOUSLY.

Then can you provide the math to show there is disagreement at bit
position (1,1) for me?

It's DEFINED to disagree at (1,1), DUMBASS.
AND at (2,2). AND at (3,3). AND at (4,4).
BY DEFINITION. The string that disagrees at all
those places DOES exist, and it's not on your list,
even though it IS in P(S).

The fact that the contradiction occurs at a
single point

IT DOES NOT, dumbass.
You're just lying.

There is a single subset, call it R, which causes the contradiction.

NO, there isn't. There are ZILLIONS of subsets not
on your list. ALL of them contradict your argument.
The anti-diagonal proof just happens to construct one of them
specifically.

And I claim there
are subsets of the diagonal set , which do not contain R, and so do not
cause a contradiction.

That doesn't change the fact that R still exists and R still
causes the contradiction (ALONG with a zillion OTHER partners
in that crime).

The "?" in the matrix above is marking the
subset R.

No, DIP***, it is NOT marking that.
It is marking whether "2" is or isn't in the set.
Each individual bit-position in the bit-string marks
whether ONE element of S (or N) is (or isn't) in the
overall subset represented by the bit-string.

.

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