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

On 16.01.2025 23:22, Jim Burns wrote:

On 1/15/2025 1:13 PM, WM wrote:

On 14.01.2025 19:41, Jim Burns wrote:

On 1/14/2025 4:07 AM, WM wrote:

On 13.01.2025 20:31, Jim Burns wrote:

 

A step is never from finite to infinite.

>
The dark realm is appears infinite.

 Nowhere,
among what appears and
among what doesn't appear,
is there a step from finite to infinite.

There is a step between the largest number accessed in the system by FISONs and its successors, which are dark but may become visible.
But the border is not fixed.
There is a step between the largest number accessible in a system and its successors, which are dark and remain dark.
But the border is not sharp
An example is the pocket calculator where all numbers after 9.999999999*10^99 are dark and remain dark. But already many smaller numbers are dark and remain dark.
The domain of visible numbers ends at 10^10.

Nowhere,
among what appears and
among what doesn't appear,
is there finite ω-1 and infinite (ω-1)+1

So it appears because ω and ω-1 are dark. But if ω is assumed to exist, then there is a set cotaining ω elements. From this set one element can be subtracted.

 Also, more generally,
there is no infiniteˢᵉᵗ smaller than ℕ

Therefore ℕ \ {1} is finite. But it appears infinite like all sets which cannot be counted by FISONs. All dark numbers larger than ω/n appear infinite.

 ⎛ For each finiteᵒʳᵈ k
⎜ ∃fₖ one.to.one: ⟦0,k⦆ ⇉ 𝕌

There is no one-to-one outside of the visible domain (except that identity mappings are assumed to exist).

None of those finitesᵒʳᵈ is the size of ℕ

All endsegments following the first sets below
{1} | {2, 3, 4, ...}
{1, 2} | {3, 4, ...}
{1, 2, 3} | {4, ...}
...
appear infinite. The first sets beyond the visible domain of FISONs appear infinite but are finite. Only when the endsegments have lost, one by one, all elements, the first set is actually infinite.
Regards, WM