Re: 2N=E

On 11/2/2024 5:34 PM, WM wrote:

On 02.11.2024 21:53, Jim Burns wrote:

On 11/2/2024 1:24 PM, WM wrote:

If all multiples of 2 smaller than ω are doubled,
then this doubling results in larger numbers than doubled.

>
If n is finite  ∧  ω ≤ n+n
then ω is finite.

>
That might appear so in a set without dark numbers.

For anything, for any φ,
if n is finite  ∧  φ ≤ n+n
then φ is finite.
Each subset Aₙ of ⟦0,n⟧ is two.ended.or.{}
⟦n,n+n⟧ = n+ᵉᵃᶜʰ ⟦0,n⟧
For each subset Bₙ₊ₙ ⊆ ⟦n,n+n⟧
Bₙ₊ₙ = n+ᵉᵃᶜʰ Aₙ ⊆ ⟦0,n⟧
Aₙ is two.ended.or.{}
  If i,f are two ends of Aₙ
  then n+i,n+f are two ends of Bₙ₊ₙ
Each subset Bₙ₊ₙ of ⟦n,n+n⟧ is two.ended.or.{}
Consider ⟦0,φ⟧ ⊆ ⟦0,n+n⟧
For each subset Cᵩ ⊆ ⟦0,φ⟧
Cᵩ = Aₙ∪Bₙ₊ₙ
Aₙ ⊆ ⟦0,n⟧  ∧  Bₙ₊ₙ ⊆ ⟦n,n+n⟧
Aₙ and Bₙ₊ₙ are each two.ended.or.{}
If Aₙ = {} = Bₙ₊ₙ  then Cᵩ = {}
If Aₙ ≠ {} = Bₙ₊ₙ  then Cᵩ = Aₙ is two.ended
If Aₙ = {} ≠ Bₙ₊ₙ  then Cᵩ = Bₙ₊ₙ is two ended
If Aₙ ≠ {} ≠ Bₙ₊ₙ  then
min.Cᵩ = min.Aₙ
max.Cᵩ = max.Bₙ₊ₙ
and Cᵩ is two.ended
Each subset Cᵩ of ⟦0,φ⟧ is two.ended.or.{}
Therefore,
if n is finite
   (each subset Aₙ of ⟦0,n⟧ is two.ended.or.{})
and φ ≤ n+n
   (⟦0,φ⟧ ⊆ ⟦0,n+n⟧)
then φ is finite
   (each subset Cᵩ of ⟦0,φ⟧ is two.ended.or.{})
In particular,
if n is finite  and  ω ≤ n+n
then ω is finite.

It is not true when dark numbers come into play.

The only numbers which have come into play here
are φ such that each subset of ⟦0,φ⟧ is two.ended.or.{}
Even ω is shown to have
each subset of ⟦0,ω⟧ is two.ended.or.{}
Of course,
the reason to show that falsehood about ω is
to show that  'n is finite ∧ ω ≤ n+n'  must be false.
You (WM) probably did not intend to say that
there are dark numbers δ such that
each subset of ⟦0,δ⟧ is two.ended.or.{}
Which of the various things you're saying are wrong?