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

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.
ω is the set of finite.ordinals.
k ∈ ⟦0,ω⦆  :⇔  finite.ordinal k
Part of what we mean by 'finite k' is that
the priors ⟦0,k+1⦆ of k+1 cannot map to
the priors ⟦0,k⦆ of k
¬∃¹ᵗᵒ¹f:⟦0,k+1⦆⇉⟦0,k⦆
It is a theorem[1] that
⎛ if ¬∃¹ᵗᵒ¹f:⟦0,k+1⦆⇉⟦0,k⦆
⎝ then ¬∃¹ᵗᵒ¹g:⟦0,k+2⦆⇉⟦0,k+1⦆
Which means
⎛ if k is finite
⎝ then k+1 is finite
[1]
⎛ If ∃¹ᵗᵒ¹g:⟦0,k+2⦆⇉⟦0,k+1⦆
⎜ then ∃¹ᵗᵒ¹f:⟦0,k+1⦆⇉⟦0,k⦆
⎜⎛ Define
⎜⎜ f(g⁻¹(k+1)) = g(k+2)
⎜⎝ otherwise f(j) = g(j)
⎜
⎜ Contrapositively,
⎜ if ¬∃¹ᵗᵒ¹f:⟦0,k+1⦆⇉⟦0,k⦆
⎝ then ¬∃¹ᵗᵒ¹g:⟦0,k+2⦆⇉⟦0,k+1⦆
Therefore,
k ∈ ⟦0,ω⦆  ⇒  k+1 ∈ ⟦0,ω⦆
Consider the after.segments of the finite.ordinals.
Define
Eᑉ(n) = {n+1, n+2, ...} ⊆ ⟦0,ω⦆
Eᑉ(0) = {1, 2, 3, ...} ⊆ ⟦0,ω⦆
Eᑉ(1) = {2, 3, 4, ...} ⊆ ⟦0,ω⦆
Eᑉ(2) = {3, 4, 5, ...} ⊆ ⟦0,ω⦆
...
⎛ Assume Eᑉ(ω-1) = {}
⎜
⎜ ∀k ∈ N: Eᑉ(n) = {n+1}∪Eᑉ(n+1)
⎜ ω-1 ∈ Eᑉ(ω-2)
⎜
⎜ k ∈ ⟦0,ω⦆  ⇒  k+1 ∈ ⟦0,ω⦆
⎜ ω-1 ∈ Eᑉ(ω-2)  ⇒
⎜ (ω-1)+1 ∈ Eᑉ(ω-2)  ⇒
⎜ (ω-1)+2 ∈ Eᑉ(ω-2)  ⇒
⎜ (ω-1)+3 ∈ Eᑉ(ω-2)  ⇒
⎜ ...
⎜
⎜ Eᑉ(ω-2) = {ω-1, (ω-1)+1, (ω-1)+2, ...}
⎜ Eᑉ(ω-1) = {(ω-1)+1, (ω-1)+2, ...} ≠ {}
⎝ Contradiction.
No last finite ordinal ω-1 exists.
Its existence gives us the contradictions.
No end.segment E(ω-1) exists.
Its existence gives us the contradictions.

Each set has
its emptier.by.one and fuller.by.one counterparts.

>
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.

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.
That contradicts |⟦k,ω⦆| being finite.
No _finite_ cardinal j = |⟦k,ω⦆| which doesn't
give us the contradictions  exists.