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

Chris M. Thomasson used his keyboard to write :

On 12/4/2024 10:22 AM, FromTheRafters wrote:

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.

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

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

>
Conclusion not supported by facts.

>
Is it "pair-wise" inclusion, or "super-task" inclusion?
>
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.

>
{ 1 - 1, 2 - 1, 3 - 1, ... } = { 0, 1, 2, ... }
>
{ 0 + 1, 1 + 1, 2 + 1, ... } = { 1, 2, 3, ... }
>
A direct mapping between them?

Yes, which more than just hints at a bijection. A bijection doesn't care about the symbols, only some idea of 'same size' or 'just as many'. An intersection requires knowing what symbols are in each set in order to 'find' matches. His infinite intersection of all endsegment sets is doomed to failure in the first iteration.