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

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