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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Sat, 21 Dec 2024 22:58:36 +0100
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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