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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 12:34:14 +0100
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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?

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. Therefore the sequence of 
endsegments cannot become empty (i.e., not all natnumbers can be applied 
as indices) unless the empty endsegment is reached, and before finite 
endsegments, endsegments containing only 1, 2, 3, or n ∈ ℕ numbers, have 
been passed.

These however, if existing at all, cannot be seen. They are dark.
>> That means all numbers are lost by loss of
>> one number per term.
>>
>> That implies finite endsegments.
> 
> No.
> Yes, each number is lost by loss of
> one number per term.
> However,
> each end.segment is not finite.

If all natnumbers become indices, they all have left the endsegments. 
Then the last endsegment is empty. Otherwise not all natnumbers  can 
become indices.

>> As long as
>> any natural number remains,
>> the sequence of lost numbers is finite
>> (ended by the remaining number).
> 
> Claims P⇒Q and ¬Q⇒¬P are both.true or both.false.
> 
> P⇒Q above, ¬Q⇒¬P below.
> 
> As long as
> the sequence of lost numbers is infinite,
> no natural number remains.
> 
> I agree.

"As long as" is bewildering here. The sequence of lost numbers is 
infinite if no endsegment contains a number. But there is only one 
endsegment containing no number.

Regards, WM