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

On 1/13/25 7:42 AM, WM wrote:

On 13.01.2025 03:54, Richard Damon wrote:

On 1/12/25 5:58 AM, WM wrote:

On 11.01.2025 14:34, joes wrote:
>

1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2, ω+4, ..., ω2.

No. There is no x e N such that 2*x >= omega. You have listed two
consecutive infinities on the right.

>
There is a basic law: When a sequence of regular distances is multiplied by 2, then a sequence of regular distances results.

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Which isn't applicable, since their isn't such a "regular distance"

 Between all natural numbers, there is a regular distance. When doubled, the new ones do not fit between the old ones and ω because nothing fits between ℕ and ω.

But omega isn't a natural number, so the space between the ... and omega isn't the same as the space between two consecutive natural numbers.
And the distance between the ... and omega *IS* big enough to fit the doubling of all the numbers (without needing to make any new ones).

 

your starting premise is just

 to double all numbers which fit between 1 and ω.

Which are still just the even numbers between 1 and omega.

 

The interval (0, ω)*2 becomes (0, ω*2) with ω in the middle. Below ω the newly created even numbers cannot be inserted, because more than all even natural numbers do not exist in actual infinity.
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Every contrary opinion is based on potential infinity.
>

>
Nope, your "opinion" is just based on your

 correct understanding that all between 1 and ω is available for doubling and nothing can be inserted.

And nothing NEEDS to be inserted.
You still are stuck on the fact that no new natural number needed to be created, because we started with an unbounded set of them, so there was always more.
That is a concept your "naive finite logic" just can't handle.

 Regards, WM