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

On 10.01.2025 19:24, joes wrote:

Am Fri, 10 Jan 2025 18:33:31 +0100 schrieb WM:

On 10.01.2025 13:41, Richard Damon wrote:
>

"Nubmers" or "Sets" don't evolve.

They either are members, or not.
 

Fine. Then the set of natural numbers is completed. Multiply all natural
numbers by 2. The set of even numbers then doubles.

Very wrong. The set {2*k for k e N} = G = 2N = {2*1, 2*2, 2*3, ...} =
{2, 4, 6, ...} is a proper subset of N.

Just this very subset is complete. That means no further element can be added.
 > I have no idea what you think.
I see. Otherwise you would not claim completeness and variabiliyt at the same time.

The domain below ω
is unable to absorb new numbers. What happens to the newly created even
numbers?

None are "created". The multiples of 4 are ALSO in the original set.

All elements of the original set can be removed. Then the set it empty. But if the same number of elements is added, then the set has twice the original number. Therefore it was not complete before. Completeness however was the premise.

{1, 2, 3, 4, 5, ..., n} contains roughly twice the even numbers.

Could you formalise what you mean here?

The ration |{1, 2, 3, ..., 2n}|/|{1, 2, 3, ..., n}| = 2 for all n.

 

This holds for all n.  Hence it holds for the infinite set.

No. The limit of the ratio is different from the ratio of the limit.

I do not apply any limit. I apply simply all natural numbers. Note that all are applied in Cantor's bijections. No limits!
But if I apply a limit, then it is the limit of the ratio.
There is no ratio of the limit.
Regards, WM