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


Scott wrote:
I am not constrained by how many elements are in the base set.

YES, YOU ARE.

In the infinite case you are, anyway.
YOU CONSTRAINED *YOURSELF*.
YOU promised that the list would be SQUARE,
NOT that it would be COMPLETE: YOU said
that the number of rows in the list, the number of subsets,
WAS THE SAME AS the number of elements in the
base set, that they were BOTH denumerable, that they
were BOTH countably infinite. YOU SAID that your
list would have the SAME number of rows as columns
(countably-infinite, the number of finite natural numbers
in all of N).

WHEN WE EXTEND that to the finite case,
we have to KEEP it square. If you have 5 elements
in the set then you WILL have *FIVE* subsets, for
purposes of OUR diagonal theorem.

In the finite case, YOU DON'T GET to set
any constraints. It's OUR diagonal theorem
and WE are the ones who TOLD YOU that the
list had to be square. Obviously you DON'T GET
a diagonal if it isn't.

My original enumeration stands and its pretty apparent you could not
come up with the next number.

If your original enumeration was infinitely long then
nobody can WRITE OUT the missing number, any more
than you cuold write out any infinite member of the list.

I win.

No, you don't.
For EVERY SQUARE list you present, you lose.

.

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