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

On 12.01.2025 15:40, Jim Burns wrote:

On 1/10/2025 4:48 PM, WM wrote:

Elsethread:
<WM>

(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)

</WM>[1]
 It sounds as though
the only explanation which you (WM) accept
for the constancy of end.segment.size is
that elements are inserted (at the darkᵂᴹ end?)
as other elements are deleted (at the visibleᵂᴹ end?)

If only elements are deleted, then the endsegments get empty.

 (Somehow this happens. Perhaps ℕ has homeostasis.)

It does not happen. It would be necessary if all endsegments were infinite.

Infiniteⁿᵒᵗᐧᵂᴹ sets do not change.

When all numbers become indices, then they are not content. That is true in every theory.

You are inconsistent.
You claim that
all natural numbers are an invariable set.

 

But when all elements are doubled
then your set grows, showing it is not invariable.
That is nonsense.

 Perhaps this argument won't look like nonsense.
 It features an utterly.familiar property,
being.finite,

No, it depends on completeness. If all natural numbers are there such than none can be added, then doubling all of them deletes odd numbers and must create new even numbers which cannot be natural numbers.

⎜ ℕ is the set of finite ordinals.

Of all. None can be added. If all are doubled, then 50 % odd nubers are deleted, 50 % even numbers are added. Because the total sum remains constant. x = x/2 + y.

Bob cannot become absent from a finite set

Lossless exchange remains lossless in all cases. That is my premise.
And another premises is this: Regular distances in (0, ω) multiplied by 2 remain regular distances in (0, 2ω).
Regards, WM