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

On 13.01.2025 21:55, Chris M. Thomasson wrote:

On 1/13/2025 9:17 AM, WM wrote:

On 13.01.2025 17:33, Jim Burns wrote:

On 1/12/2025 2:39 PM, WM wrote:

On 12.01.2025 20:33, Jim Burns wrote:

On 1/12/2025 10:54 AM, WM wrote:

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No, it depends on completeness.

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It is completely true
that each natural number is a natural number   and
that only natural numbers are natural numbers.

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and that nothing fits between them and ω.

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Yes.

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Therefore doubling of all natural numbers creates numbers larger than ω.

 No. double all of the natural numbers:
 { 1 * 2, 2 * 2, 3 * 2, 4 * 2, ... } = { 2, 4, 6, 8, ... }
 All of those results are natural numbers. They were already there...
 

That is wrong in all kinds of infinity.
∀n ∈ ℕ: 2n > n. All numbers are doubled. Their number remains the same (not only their cardinality, but their reality). Half of all, the odd numbers, are deleted. Half must be new.
In potential infinity they simply are added by expanding the sequence because there is no ω. I actual infinity there is no space available below ω.

Doubling of all n deletes the odd numbers but cannot change the number of numbers, therefore creates even numbers. They do not fit below ω. Remember: nothing fits between them and ω.

Regards, WM