Re: how

On 4/11/2024 8:07 AM, WM wrote:

Le 11/04/2024 à 01:05, Richard Damon a écrit :

On 4/10/24 4:14 PM, WM wrote:

Le 10/04/2024 à 01:06, Richard Damon a écrit :

Nope, you can use ALL of the Natural Numbers.

>
You can use only a small minority because
almost all remain unused:
∀n ∈ ℕ_used: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

>
Which ones were unused by e = 2*n?

>
Those of {1, 2, 3, ...} with less than ℵo successors.

{1,2,3,…} is
the least.upper.bound of all finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k}
If not( end.of.finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k} k ∈ {1,2,3,…} )
then not( {1,2,3,…} upper.bound )
and not( {1,2,3,…} least.upper.bound )
If not.end.of.finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,δ} δ ∈ {1,2,3,…}
then {1,2,3,…}\{δ} lesser.upper.bound
and not ( {1,2,3,…} least.upper.bound )
If {1,2,3,…} least.upper.bound
then
k ∈ {1,2,3,…}  ⟺
k end.of.finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k}
k⁺¹ finiteⁿᵒᵗᐧᵂᴹ  if k finiteⁿᵒᵗᐧᵂᴹ.
m+k = (m+k⁻¹)⁺¹ finiteⁿᵒᵗᐧᵂᴹ  if m+k⁻¹ finiteⁿᵒᵗᐧᵂᴹ.
m⋅k = (m⋅k⁻¹)+m finiteⁿᵒᵗᐧᵂᴹ  if m⋅k⁻¹ finiteⁿᵒᵗᐧᵂᴹ.

Those of {1, 2, 3, ...} with less than ℵo successors.

For each m ∈ {1,2,3,…}:
( for each k ∈ {1,2,3,…}
m+k is a successor of m in {1,2,3,…} )
For each m ∈ {1,2,3,…}:
( |{m+1,m+2,m+3,…}| = |{1,2,3,…}| = ℵ₀ )