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

On 21.12.2024 20:32, Jim Burns wrote:

On 12/21/2024 6:34 AM, WM wrote:

On 20.12.2024 19:48, Jim Burns wrote:

On 12/19/2024 4:37 PM, WM wrote:

 

That means all numbers are lost by loss of
one number per term.
>
That implies finite endsegments.

>
Q. What does 'finite' mean?

Finite endsegments have a natural number of elements.

 Consider end.segments of the finite cardinals.
 Q. What does 'finite' mean,
'finite', whether darkᵂᴹ or visibleᵂᴹ?

Finite endsegments cannot be visible

 

Here is a new and better definition of endsegments
>
E(n) = {n+1, n+2, n+3, ...} with E(0) = ℕ.
>
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}
means that the sequence of endsegments can decrease only by one natnumber per step.

 E(n+1) is larger.than each of
the sets for which there are smaller.by.one sets.
E(n+1) isn't any of
the sets for which there are smaller.by.one sets.

No. E(n+2) is smaller than E(n+1) by one element, namely n+2.

 E(n+1) isn't smaller.by.one than E(n).
E(n+1) is emptier.by.one than E(n)

It is also smaller, but we cannot distinguish ℵ₀ and ℵ₀ - 1.
{1, 2, 3, ..., ω} is smaller by one element than {0, 1, 2, 3, ..., ω}.
|{1, 2, 3, ..., ω}| < |{0, 1, 2, 3, ..., ω}|

 

Therefore the sequence of endsegments
cannot become empty

 Yes, because
the sequence of end.segments
can become emptier.one.by.one, but
it cannot become smaller.one.by.one.

Both happens. Cantor's bijections are nonsense.

 

(i.e., not all natnumbers can be applied as indices)

 Each finite.cardinal can be applied,
which makes the sequence emptier.by.one
but does not make the sequence smaller.by.one.

Rest deleted because it is wrong. There are |ℕ|^2 + 1 fractions.
Regards, WM