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

On 11/27/2024 6:04 AM, WM wrote:

On 26.11.2024 20:44, Jim Burns wrote:

On 11/26/2024 2:15 PM, WM wrote:

On 26.11.2024 19:49, Jim Burns wrote:

There are no last end.segments of ℕᶠⁱⁿ
There are no finitely.sized end segments of ℕᶠⁱⁿ
There are no finite cardinals common to
 each end.segment of ℕᶠⁱⁿ

set ℕᶠⁱⁿ of finite cardinals == can.change.by.1

That is a contradiction.

>
It contradicts ℕᶠⁱⁿ being finite, nothing else.

>
It contradicts inclusion monotony.
>

If there are no common numbers,
then all numbers must have been lost.
But then no numbers are remaining.

>
Yes.

>
Then also
no numbers are remaining in the endsegments.

Yes,
for each number (finite cardinal)
there is an end segment such that
the number isn't in the end segment.
However,
one makes a quantifier shift, unreliable,
to go from that to
⛔⎛ there is an end segment such that
⛔⎜ for each number (finite cardinal)
⛔⎝ the number isn't in the end segment.
The unreliability of quantifier shift is
unrelated to infinity, either ours or yours.
Yes:
for each j in {1,2,3}
there is k in {4,5,6} such that
j+3 = k
No:
⛔⎛ there is k in {4,5,6} such that
⛔⎜ for each j in {1,2,3}
⛔⎝ j+3 = k

Each finite.cardinal k is countable.past to
  k+1 which indexes
   Eᶠⁱⁿ(k+1) which doesn't hold
    k which is not common to
     all end segments.
>
Each finite.cardinal k is not.in
  the intersection of all end segments,
  the set of elements common to all end.segments,
   which is empty.
>
No numbers are remaining.

>
That is true.
But you claimed that every endsegment is infinite.

Yes,
if an end.segment is the intersection of all,
then it is the last end segment of all.
However,
there are no last end.segments of ℕᶠⁱⁿ
There is no intersection.end.segment.
Each end.segment is infinite.
Their intersection of all is empty.
These claims do not conflict.

In an infinite endsegment
numbers are remaining.
In many infinite endsegments infinitely many numbers are the same.

And the intersection of all,
which isn't any end.segment,
is empty.

Then there are finite endsegments because
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.

>
For each cardinal.which.can.change.by.1 j
|Eᶠⁱⁿ(k)| is larger than j
|Eᶠⁱⁿ(k)| isn't j
 |Eᶠⁱⁿ(k)| isn't any cardinal.which.can.change.by.1
|Eᶠⁱⁿ(k)| cannot change by 1
|Eᶠⁱⁿ(k+1)| = |Eᶠⁱⁿ(k)|

A cardinal.which.can.change.by.1 is finite.
No end.segment of ℕᶠⁱⁿ is finite.

That does not contradict the fact that
infinite endsegments have infinitely many numbers in common

And aren't the intersection.end.segment.

and hence an infinite intersection.

Each finite.cardinal k is countable.past to
  k+1 which indexes
   Eᶠⁱⁿ(k+1) which doesn't hold
    k which is not common to
     all end segments.
Each finite.cardinal k is not.in
  the intersection of all end segments,
  the set of elements common to all end.segments,
   which is empty.