Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

WM <wolfgang.mueckenheim@tha.de> wrote:

On 26.11.2024 18:24, Richard Damon wrote:

WM <wolfgang.mueckenheim@tha.de> wrote:

 

And nothing in the logic says that white hats go away.

 
Nothing in logic allows that. But Cantor claims it erroneously.

 

Cantor NEVER had the white hats go away, he just showed you could match
every number with a black hat

 
That means all white hats must disappear under black hats. That means
the white and black must be exchanged until no white remains. That is
impossible.
 
Regards, WM
 

Why is there a white hat under the black hat?

They are two *DIFFERENT* sets, one with the number {1, 2, 3, 4, …} each
with a white hat, and one with the numbers {10, 20, 30, 40, …} each with a
black hat. Just because we gave 10 a white hat in the first set doesn’t
give the 10 in the second set that hat, then are DIFFERENT sets.

We can exchange the white hat from the first set on 1 with the black hat on
the second set on 10.
We then swap hats between 1st set 2 and 2nd set 20, and so on for the first
set n and the second set 10n, and we see that ALL the numbers in the first
set get their black hats and all the numbers in the second set get white
hats, and thus we have proven that this is a bijection, and the sets use
have the same cardinality.

You are just showing you don’t understand what is happening, because you
just don’t understand how mathematics or logic actually work but are just
following some made up rules that just blow up in your face when you
mishandle them.