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

On 19.12.2024 19:29, Jim Burns wrote:

On 12/19/2024 9:50 AM, WM wrote

⋂{E(1),E(2),...} is empty.
Other numbers are not in the argument.

>
But E(1) is full.
The only way to get rid of content is
to proceed by
∀k ∈ ℕ:
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k},
i.e. to lose one number per term of
the function
f(k) = ∩{E(1), E(2), ..., E(k)}.

 f(k) = E(k) = ⟦k,ℵ₀⦆
 Yes,
after infinitely.many one.number.losses,
infinitely.many numbers are lost.

That means all numbers are lost by loss of one number per term. That implies finite endsegments.

 However,
each number is not.after infinitely.many numbers,
each loss is not.after infinitely.many losses.

As long as any natural number remains, the sequence of lost numbers is finite (ended by the remaining number).

 

The empty term

 There is no darkᵂᴹ or visibleᵂᴹ empty term.

Finite intersections of endsegments preceding the empty intersection of all endsegments are dark.

 ⎛ Assume otherwise.

For my argument I don't need empty endsegments.

The empty term
has lost all natnumbers one by one.

 One.by.one and (darkᵂᴹ or visibleᵂᴹ) endlessly.

If Cantor uses all indices for his bijections the intersection of all endsegments must be empty.
Regards, WM