Re: how

On 6/5/2024 6:43 AM, WM wrote:

Le 04/06/2024 à 23:31, Jim Burns a écrit :

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.

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

>
Of course.

Assumption 2 in detail.
For numbers i,j,k countable.to by.1 from.0
i before j before k  implies  i before k
exactly one is true of
'j before k', 'j after k', 'j equals k'
i < j < k  ⇒  i < k
j <≠≯ k  ∨  j ≮=≯ k  ∨  j ≮≠> k
For number k countable.to by.1 from.0
exists set {k≥} of
numbers up.to.k countable.to by.1 from.0
such that,
for each foresplit F of {k≥}
exists i last.in.F  and
exists j first.in.{k≥}\F
∀F ⊆ {k≥}: ∅ ≠ F ᴬ<ᴬ {k≥}\F ≠ ∅  ⇒
∃i ∈ F: F ᴬ≥ i  ∧
∃j ∈ {k≥}\F: j ≥ᴬ {k≥}\F
Notation:
i.successor j = i⁺¹
j.predecessor i = j⁻¹
Each countable.to has a successor.
Each nonzero countable.to has a predecessor.

But that is a small minority

All of ℕ⁺ is in ℕ⁺
Nothing but ℕ⁺ is in ℕ⁺

because you cannot count all:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
Whatever you count, most numbers remain uncounted.

Avogadroᴬᵛᵒᵍᵃᵈʳᵒ remains uncounted in
the universe I inhabit.
However,
the set {Avogadroᴬᵛᵒᵍᵃᵈʳᵒ≥} exists as described
with last.before and first.after each split.
Thus,
Avogadroᴬᵛᵒᵍᵃᵈʳᵒ is countable.to by.1 from.0
and also uncounted in the universe I inhabit.
(Avogadro := 6.02214076×10²³)

ℵo is much larger than
the collection |ℕ_def| of all counted n.

ℵ₀ is defined to be |ℕ⁺| of all
numbers countable.to by.1 from.0
Defined to be.
If 3 was defined to be the number of corners of a triangle,
and I said, "But wait! There are more/fewer corners!"
either I didn't know what a triangle is
or I didn't know what 3 is.
ℵ₀ is defined to be |ℕ⁺|
If ℕ_def isn't ℕ⁺
and |ℕ_def| isn't |ℕ⁺|
then what are you (WM) objecting to?

We agree if
darkᵂᴹ numbers are countable.to by.1 from.0.

>
They remain uncounted.

Each number in ℕ⁺ is countable.to by.1 from.0
That's how ℕ⁺ is defined.
If darkᵂᴹ numbers are in ℕ⁺
then they are countable.to by.1 from.0
If not, then not.

Proof:
For all counted n:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Don't you believe this statement?
Or don't you understand it?

ℕ\{1,2,3,...,n}  =  ℕ⁺\{n≥}
ℕ⁺ is the set of countable.to by.1 from 0
ℵ₀ is defined to be |ℕ⁺|
For each j ∈ ℕ⁺
exists unique k ∈ ℕ⁺
k = j⁺¹ > j
1.to.1 j ⟼ j⁺¹:  ¬∃j₂≠j: j₂⁺¹=j⁺¹
For each j ∈ ℕ⁺\{i≥}
exists unique k ∈ ℕ⁺\{i⁺¹≥}
k = j⁺¹ > j
1.to.1 j ⟼ j⁺¹:  ¬∃j₂≠j: j₂⁺¹=j⁺¹
|ℕ⁺\{i≥}| ≤ |ℕ⁺\{i⁺¹≥}|
  ℕ⁺\{i≥}  ⊇  ℕ⁺\{i⁺¹≥}
|ℕ⁺\{i≥}| ≥ |ℕ⁺\{i⁺¹≥}|
|ℕ⁺\{i≥}| = |ℕ⁺\{i⁺¹≥}|
not.exists n ∈ ℕ⁺:  |ℕ⁺\{n≥}| ≠ ℵ₀
| Assume otherwise.
| Assume nₓ ∈ ℕ⁺:  |ℕ⁺\{nₓ≥}| ≠ ℵ₀
|
| nₓ ∈ ℕ⁺
| exists {nₓ≥} as described above
|
| Define
| Fₓ = {j∈{nₓ≥}|∀i≤j:|ℕ⁺\{i≥}|=ℵ₀}
| Fₓ is a foresplit of {nₓ≥}
|
| Because {nₓ≥} is as described above
| exists iₓ last.in.Fₓ  and
| exists jₓ first.in.{nₓ≥}\Fₓ
| We know, using the definition of Fₓ
| |ℕ⁺\{iₓ≥}| = ℵ₀
| |ℕ⁺\{jₓ≥}| ≠ ℵ₀
| |ℕ⁺\{iₓ≥}| ≠ |ℕ⁺\{jₓ≥}|
|
| However,
| jₓ = iₓ⁺¹
| From above
| |ℕ⁺\{iₓ≥}| = |ℕ⁺\{iₓ⁺¹≥}|
| Contradiction.
Therefore,
not.exists n ∈ ℕ⁺:  |ℕ⁺\{n≥}| ≠ ℵ₀

For all counted n:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.

not.exists n ∈ ℕ⁺:  |ℕ⁺\{n≥}| ≠ ℵ₀
What do you (WM) think that proves?