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

On 14.12.2024 05:54, Jim Burns wrote:

On 12/13/2024 2:31 PM, WM wrote:

On 13.12.2024 20:00, Jim Burns wrote:

On 12/13/2024 6:25 AM, WM wrote:

 

Ignoring that Cantor's claim requires to
empty the endsegments from all natural numbers
in order to use them as indices in mappings

>
Each finite.cardinal
⎛ is first in an end.segment.
⎝ is used as a index of an end.segment.
>
Each finite.cardinal
⎛ is absent from an end segment.
⎝ is emptied from the end.segments.
>
Require at will, sir.

>
If endegments were defined as
E(n) = {n+1, n+2, ...}:
>
E(0) = {1, 2, 3, ...}
E(1) = {2, 3, 4, ...}
E(2) = {3, 4, 5, ...}
...
E(ω-1} = { }.
Then this change from content to index
would even be more obvious.

 The index of Eᑉ(2) is content of Eᑉ(1), etc.
The number doesn't change.
Which after.segment changes.
 ----
One problem which
Eᑉ(ω-1) = {}
has  is that
'finite' is NOT defined the way in which
you (WM) think 'finite' should be,
which means
ω is NOT defined the way in which
you (WM) think ω should be.

Don't say what not is. Explain your vision of the problem:
If ℕ is a set, i.e. if it is complete such that all numbers can be used for indexing sequences or in other mappings, then it can also be exhausted such that no element remains. Then the set of what remains unused, i.e., of intersections of endsegments
(1)   E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content. Then, by the law
(2)   ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
the content must become finite.

Which means
⎛ if k is finite
⎝ then k+1 is finite

This contradicts the fact that nothing remains but every element can go only as a single.

Therefore,
k ∈ ⟦0,ω⦆  ⇒  k+1 ∈ ⟦0,ω⦆

This contradicts the fact that nothing remains but every element can go only as a single. Which if the two conditions (1) and (2) has to be sacrificed?

If all natnumbers can be used for mappings as indices
then every natnumber has to leave the sequence of endsegments.
Do you agree?

 Yes.
 

Therefore their intersection is empty.
Do you agree?

 Yes.
 

Emptying by one only implies
finite endsegment intersetcions.
Do you agree?

 No.

(1) is wrong?

 

If not describe the process according to your opinion.

 Each end.segment ⟦k,ω⦆ of ⟦0,ω⦆ contains,
for each finite.cardinal j
a subset ⟦k,k+j⟧ holding more.than.j.many.

Then the intersection is never empty (or only for definable finite cardinals)?

 That contradicts |⟦k,ω⦆| being finite.

Otherwise (1) /\ (2) is conztradicted.
Regards, WM