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

Ross Finlayson laid this down on his screen :

On 12/04/2024 02:33 AM, FromTheRafters wrote:

WM formulated the question :

On 03.12.2024 21:34, Jim Burns wrote:

On 12/3/2024 8:02 AM, WM wrote:

>

E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term
have identical limits.

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An empty intersection does not require
  an empty end.segment.

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A set of non-empty endsegments has a non-empty intersection. The
reason is inclusion-monotony.

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Conclusion not supported by facts.

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Is it "pair-wise" inclusion, or "super-task" inclusion?
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Which inclusion is of this conclusion?
>
They differ, ....

I like to look at it as {0,1,2,...} has a larger 'scope' of natural numbers than {1,2,3,...} while retaining the same set size. It does this by not being finite. In the first one, zero is included but in the second and further ones zero 'cannot' be in any further of the infinitely many intersections of endsegments. No need to 'supertask' or 'pair' anything at all, it's already a done deal.
Of course, sometimes these things are not obvious but hidden in the details.