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


Scott wrote:
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".

It's still a square.

OK, fine, it's an infinity x infinity square.

Which means you could not come up with the number.

You're a damn liar. I told you, AND CANTOR TOLD YOU
A DOZEN DECADES AGO, the number is:
the bit-string whose nth bit is 1 minus the nth bit
of the nth row. DONE.

That applies REGARDLESS OF THE SIZE OF THE LIST,
as long as it is square. It works 69 subsets of a set of size 69.
It works for 2 subsets of a set of size 2. It works for 4 subsets
of a set of size 4. And as Cantor has proved to you dozens of
times by now, it works for denumerably many subsets of a
denumerable set. THE PROOF IS THE SAME. The proof
DOESN'T EVEN MENTION the size of the set!

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.

Oh come on, you've got to have a brain cell in that head somewhere.

The "open" slot is for your number D.

NO, it isn't! NOTHING with "open" even COUNTS as
the list YOU started with! YOU said YOU could come up with
a list of all the subsets! Everything on your list has NO opens in it!
It ONLY has 0's and 1's!

Slot 3 is where the enumeration of D is!

There is NO SUCH THING as "the enumeration of D", idiot!
What you MEANT to say was "Slot 3 is where D is" and that the
enumeration maps D to 3. BUT IT DOESN'T. D of YOUR original
list is NOT ON your list because whatEVER your list has at position
3,3, * D HAS THE OPPOSITE *. So row 3 of your list is NOT D.
NO row of your list is D.

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