Re: how

On 6/6/2024 9:27 AM, WM wrote:

Le 05/06/2024 à 23:31, Jim Burns a écrit :

Assumption (2.) describes
objects in our familiar arithmetic.

>
That is true

>
Thank you.

>

Assumption 2.
ℕ⁺ holds all.and.only
numbers countable.to by.1 from.0

>
Of course.

Darkᵂᴹ numbers do not exist in ℕ⁺

>
That is potential infinity.

They are objects in our familiar arithmetic.

It is not Cantor's actual infinity.
∀n ∈ ℕ⁺: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Define
{n<}  =  ℕ⁺\{1,2,3,...,n}
n.followers.in.ℕ⁺
|n<|  =  |{n<}|  =  |ℕ⁺\{1,2,3,...,n}|
countable.to n is ℵ₀.followed.in.ℕ⁺  ⇔
|ℕ⁺\{1,2,3,...,n}| = ℵ₀  ⇔
ℵ₀=|n<|
Each k ∈ ℕ⁺ is ℵ₀.followed.in.ℕ⁺
| Assume otherwise.
| Assume k ∈ ℕ⁺ is not ℵ₀.followed.in.ℕ⁺
| ¬(ℵ₀=|k<|)
|
| exists first j ∈ ℕ⁺ such that
| j is not ℵ₀.followed.in.ℕ⁺ and
| each i < j is ℵ₀.followed.in.ℕ⁺
|
| j is not ℵ₀.followed.in.ℕ⁺ and
| j-1 is ℵ₀.followed.in.ℕ⁺
| |j<| ≠ |j-1<|
|
| However,
| i ⟼ i+1  is 1.to.1 from.ℕ⁺ to.ℕ⁺
| i ⟼ i+1  is 1.to.1 from.{j-1<} to.{j<}
| |j-1<|  ≤  |j<|
|
| {j-1<}  ⊇  {j<}
| |j-1<|  ≥  |j<|
| |j-1<|  =  |j<|
| Contradiction.
Therefore,
each k ∈ ℕ⁺ is ℵ₀.followed.in.ℕ⁺

It is not Cantor's actual infinity.

It is our familiar arithmetic.