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

On 12/21/2024 6:34 AM, WM wrote:

On 20.12.2024 19:48, Jim Burns wrote:

On 12/19/2024 4:37 PM, WM wrote:

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

>
Q. What does 'finite' mean?

Consider end.segments of the finite cardinals.
Q. What does 'finite' mean,
'finite', whether darkᵂᴹ or visibleᵂᴹ?

Here is a new and better definition of endsegments
>
E(n) = {n+1, n+2, n+3, ...} with E(0) = ℕ.
>
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}
means that the sequence of endsegments can decrease only by one natnumber per step.

E(n+1) is larger.than each of
the sets for which there are smaller.by.one sets.
E(n+1) isn't any of
the sets for which there are smaller.by.one sets.
E(n+1) isn't smaller.by.one than E(n).
E(n+1) is emptier.by.one than E(n)

Therefore the sequence of endsegments
cannot become empty

Yes, because
the sequence of end.segments
can become emptier.one.by.one, but
it cannot become smaller.one.by.one.

(i.e., not all natnumbers can be applied as indices)

Each finite.cardinal can be applied,
which makes the sequence emptier.by.one
but does not make the sequence smaller.by.one.

unless the empty endsegment is reached,

Each end.segment is emptier.by.one.
No end.segment is smaller.than the first end.segment ℕ
The empty end.segment is not reached.
No finite.cardinal is in common with
all infinitely.many
infinite.end.segments of
finite.cardinals.
Nothing other.than a finite.cardinal is in
any end.segment of the finite.cardinals,
or in their intersection.
The intersection of all infinitely.many
infinite end.segments of finite.cardinals
is not an end.segment
but is empty.
Q. What does 'finite' mean?

unless the empty endsegment is reached,

The empty end.segment, not.existing, is not.reached.
The intersection.of.finitely.many is not.empty.
The intersection.of.all is empty.

and
before finite endsegments,
endsegments containing only 1, 2, 3, or n ∈ ℕ numbers,
have been passed.

⎛ Assume end.segment E(n) of the finite.cardinals
⎜ holds only finite.cardinal.k.many finite.cardinals.
⎜
⎜ k.sized E(n) holds
⎜ kᵗʰ.smallest, 1ˢᵗ.largest finite.cardinal E(n)[k]
⎜
⎜ If a finite.cardinal larger.than E(n)[k] exists,
⎜ it would also be in end.segment E(n) and
⎜ larger.than 1ˢᵗ.largest E(n)[k]: a contradiction.
⎜
⎜ Thus,
⎜ a finite.cardinal larger.than E(n)[k] doesn't exist.
⎜
⎜ However,
⎜⎛ for each finite.cardinal j,
⎜⎝ larger.than.j finite.cardinal j+1 exists.
⎜
⎜ Larger.than.E(n)[k] finite.cardinal E(n)[k]+1 exists.
⎝ Contradiction.
Therefore,
end.segment E(n) of the finite.cardinals does not hold
only finite.cardinal.many finite.cardinals.
There are no finite end.segments of the finite.cardinals.
Q. What does 'finite' mean?

These however, if existing at all, cannot be seen.
They are dark.

Darknessᵂᴹ and visibilityᵂᴹ don't change any of this.
There are no finite end.segments of the finite.cardinals.
We know it by the method of
assembling finite sequences of claims (proofs),
each claim of which is true.or.not.first.false (valid),
and holding those claims.we.know (theorems),
because
a finite sequence of claims,
each claim of which true.or.not.first.false,
holds only true claims.
Some claims (definitions)
we know are true because
we know how we have defined things.
Some claims (valid inferences)
we know are not.first.false because
we can inspect the finite sequence of claims.
None of _the claims_ are darkᵂᴹ,
whatever the status of _what the claims are about_
Darknessᵂᴹ or visibilityᵂᴹ of finite.cardinals
don't change _the claims_

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

>
No.
Yes, each number is lost by loss of
one number per term.
However,
each end.segment is not finite.

 

Then the last endsegment is empty.

There is no last end.segment of the finite.cardinals.
⎛ For each finite.cardinal j,
⎝ larger.than.j finite.cardinal j+1 exists.
contradicts a last end.segment.