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


Scott wrote:
George:

If the list is square, you can ALWAYS add an anti-diagonal.
NO MATTER WHAT YOU DO, YOU WILL NEVER finish
adding "all" possible anti-diagonals. If the list is both
denumerably long and denumerably wide then it ALWAYS
has an anti-diagonal. It DOES NOT MATTER how many other
anti-diagonals you did or didn't add before.
EVERY square list has an anti-diagonal that is not on it.
EVERY one. Finite, infinite, IT DOESN'T MATTER.
There is ALWAYS an anti-diagonal, just like there is always a
diagonal, and always a 1st column, and always a 69th column.

Perhaps George can put this to rest for us with a simple example/proof.
Let S = { 0, 1 }. Here is my enumeration of P(S):

1 - { }
2 - { 0 }
3 - { 1 }
4 - { 0, 1 }

DIP***: i said:
* IF THE LIST IS SQUARE *.
Read it again, dip***; YOU QUOTED IT.
Don't you even READ what YOU post??
Dip***: i said:
* Every SQUARE list has an anti-diagonal that is not on it. *

Okay, to put this to rest, show us the fifth element in the
enumeration.

Dip***: If the set only has TWO elements, then making
a SQUARE list of the subsets means you only get to put
TWO subsets on your list. And SINCE THERE ARE FOUR,
I can ALWAYS find one (The third or the fourth) that is NOT
ON YOUR LIST. The subsets are intended to be represented
by bit-strings, once you have ordered the elements in the set.
In the order that you enumerated it, YOUR list is
00
10
01
11

Which is 4x2, NOT 2x2, and NOT square.
Dumbass.


And, if you can, the sixth as well. Show us that you
cannot enumerate all elements.

But you also say if it is a square you can do it.

NO, not ALSO, but JUST in that case.

Okay, assuming an
infinite sequence of 0s and 1s, here is a square:

1 - 0000....
2 - 1000...
3 -
4 - 0100...
...

Its a 4x4 square,

NO, DIP***, THIS an infinity x infinity square;
this list is both infinitely long and infinitely wide.
You, after all, did (ignorantly) begin by saying
"assuming an infinite sequence of 0s and 1s".

I have left open slot 3 for your new number D.

No, you haven't. If this is a list of numbers then it does
NOT HAVE ANY "open" slots. "Open" is NOT a number.

.

Quantcast