Re: how

On 6/4/2024 10:10 AM, WM wrote:

Le 04/06/2024 à 04:07, Jim Burns a écrit :

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

 That is true

Thank you.

That is true
because
our familiar arithmetic is based upon
potential infinity.

a.
⎛ There is no number in ℕ⁺ larger than
⎝ all numbers in ℕ⁺
b.
⎛ There is no number in 𝔼 larger than
⎝ all numbers in 𝕆
c.
⎛ There is no number in 𝕆 larger than
⎝ all numbers in 𝔼
d.
⎛ There is no number in ℚ⁺ smaller than
⎝ all numbers in ℚ⁺
e.
⎛ There is no number in ⅟ℕ smaller than
⎝ all numbers in ⅟ℕ
f.
⎛ There is no number in ⅟ℕ smaller than
⎝ all numbers in ℚ⁺\⅟ℕ
g.
⎛ There is no number in ℚ⁺\⅟ℕ smaller than
⎝ all numbers in ⅟ℕ
h.
⎛ There is no number in ℝ⁺ smaller than
⎝ all numbers in ℝ⁺
i.
⎛ There is no number in ⅟ℕ smaller than
⎝ all numbers in ℝ⁺\⅟ℕ
j.
⎛ There is no number in ℝ⁺\⅟ℕ smaller than
⎝ all numbers in ⅟ℕ

Only in finished infinity
dark numbers are required.

However,

our familiar arithmetic is based upon
potential infinity.

Claims (a.) to (j.) are for our familiar arithmetic.
They remain correct,
whatever you call it.

For each n in ℕ⁺
n countable.to from.0
n+1 is countable.to from.n
n+1 is countable.to from.0 through.n
n+1 is in ℕ⁺
n is not larger than all numbers in ℕ⁺

>
And in particular all n have ℵo dark successors.

a.
⎛ There is no number in ℕ⁺ larger than
⎝ all numbers in ℕ⁺

And in particular all n have ℵo dark successors.

Each n ∈ ℕ⁺ has a unique successor ∈ ℕ⁺
Each follower ∈ ℕ⁺ of n ∈ ℕ⁺ has
a unique follower.successor ∈ ℕ⁺
|{m ∈ ℕ⁺: m > n}|  =  |{m ∈ ℕ⁺: m > n+1}|
There is no first n ∈ ℕ⁺: |{m ∈ ℕ⁺: m > n+1}| < |ℕ⁺|
There is no n ∈ ℕ⁺: |{m ∈ ℕ⁺: m > n+1}| < |ℕ⁺| = ℵ₀
Assumption 2.
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
We agree if
darkᵂᴹ numbers are countable.to by.1 from.0.

𝔼 is the subset of even numbers in ℕ⁺
𝕆 is the subset of odd numbers in ℕ⁺
>
⎛ There is no number in 𝔼 larger than
⎝ all numbers in 𝕆
>
⎛ There is no number in 𝕆 larger than
⎝ all numbers in 𝔼

>
Dark numbers do not unveil their mystery.

2.
Or we can assume instead that
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
ℚ⁺ holds all.and.only ratios of numbers in ℕ⁺
ℝ⁺ holds all of ℚ⁺ and all.and.only
points x between open.foresplits Fₓ and ℚ⁺\Fₓ of ℚ⁺
with no points zero distance apart
and
⅟ℕ holds all.and.only reciprocals of numbers in ℕ⁺
ℝ⁺\⅟ℕ holds all.and only the others in ℝ⁺
and
ℝ⁺ is the positive real axis.