Re: how

On 4/13/2024 8:35 AM, WM wrote:

Le 12/04/2024 à 20:57, Jim Burns a écrit :

usedᵂᴹ or not.usedᵂᴹ
not.exists
k, n in {1,2,3,…}  and
k⁺¹ not.in {1,2,3,…} or
n⁺¹ not.in {1,2,3,…} or
k+n not.in {1,2,3,…} or
k⋅n not.in {1,2,3,…}

>
All individually usable numbers satisfy
∀n ∈ ℕ_ind: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

∀n ∈ {1,2,3,…}: |{1,2,3,…}\{1,2,3,…,n}| = ℵ₀
ℵ₀ = |{1,2,3,…}|
...because
∀n ∈ {1,2,3,…}:
1.to.1 '+n': {1,2,3,…}  ⇉  {1,2,3,…}\{1,2,3,…,n}
∀n ∈ {1,2,3,…}:
|{1,2,3,…}|  ≤  |{1,2,3,…}\{1,2,3,…,n}|
|{1,2,3,…}|  ≥  |{1,2,3,…}\{1,2,3,…,n}|

All collectively useable numbers satisfy
|ℕ \ {1, 2, 3, ...}| = 0

|{1,2,3,…}\{1,2,3,…}| = 0

What elements fall between ω and ω*2?

>
ω+1 ω+2 ...

>
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.

Why 2 ?
Because ω ∈ {1,2,3,…} ?
But ω ∉ {1,2,3,…}
{0,1,2,3,…} is
the set of ordinals n such that
|⟦0,n-1⦆| ≠ |⟦0,n⦆| ≠ |⟦0,n+1⦆|
if n-1 and n+1 exist.
For example, 0-1 isn't an ordinal.
⟦0,0-1⦆ doesn't exist.
However, |⟦0,0⦆| ≠ |⟦0,0+1⦆|
⟦0,n+1⦆ = ⟦0,n⦆∪{n} exists for all ordinals
lemma.
  either
|⟦0,n-1⦆| ≠ |⟦0,n⦆| ≠ |⟦0,n+1⦆|
  or
|⟦0,n-1⦆| = |⟦0,n⦆| = |⟦0,n+1⦆|
Proof.
if ∃f: ⟦0,n+1⦆ ⇉  ⟦0,n⦆
then ∃f⤨: ⟦0,n⦆ ⇉  ⟦0,n-1⦆
for
f⤨(f⁻¹(n)) = f(n+1)
ω is the least.upper.bound of
ordinals n:  |⟦0,n-1⦆| ≠ |⟦0,n⦆| ≠ |⟦0,n+1⦆|
|⟦0,n-1⦆| ≠ |⟦0,n⦆| ≠ |⟦0,n+1⦆|  ⟸  n < ω
|⟦0,ξ-1⦆| = |⟦0,ξ⦆| = |⟦0,ξ+1⦆|  ⟸  ω < ξ
ω+1 exists
|⟦0,ω⦆| = |⟦0,ω+1⦆|  ⟸  ω < ω+1
|⟦0,ω-1⦆| = |⟦0,ω⦆| = |⟦0,ω+1⦆|
  if ω-1 exists
| Assume ω-1 exists
| ⟦0,ω-2⦆| ≠ |⟦0,ω-1⦆| ≠ |⟦0,ω⦆|  ⟸  ω-1 < ω
|
| |⟦0,ω-1⦆| ≠ |⟦0,ω⦆|
|  and
| |⟦0,ω-1⦆| = |⟦0,ω⦆|
| Contradiction.
ω-1 not.exists.
{0,1,2,3,…} is
the set of ordinals n such that
|⟦0,n-1⦆| ≠ |⟦0,n⦆| ≠ |⟦0,n+1⦆|
if n-1 and n+1 exist.
|⟦0,ω⦆| = |⟦0,ω+1⦆|
ω-1 not.exists.
ω ∉ {0,1,2,3,…}

What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?

if  |⟦0,n-1⦆| ≠ |⟦0,n⦆| ≠ |⟦0,n+1⦆|
then  |⟦0,2⋅n-1⦆| ≠ |⟦0,2⋅n⦆| ≠ |⟦0,2⋅n+1⦆|
if  n < ω
then  2⋅n < ω
∀n ∈ {0,1,2,3,…}:  n⋅2 < ω
{0,1,2,3,…}ᣔ⋅2 < ω
usedᵂᴹ or not.usedᵂᴹ
not.exists
k, n in {0,1,2,3,…}  and
k⁺¹ not.in {0,1,2,3,…} or
n⁺¹ not.in {0,1,2,3,…} or
k+n not.in {0,1,2,3,…} or
k⋅n not.in {0,1,2,3,…}
It is arithmetic.