Re: Contradiction of bijections as a measure for infinite sets

On 4/10/2024 1:57 PM, WM wrote:

Le 09/04/2024 à 20:27, Jim Burns a écrit :

On 4/9/2024 8:22 AM, WM wrote:

Le 09/04/2024 à 01:54, Jim Burns a écrit :

On 4/8/2024 9:55 AM, WM wrote:

Le 07/04/2024 à 21:47, Jim Burns a écrit :

The successor operation is closed in
the natural numbers.

>
For visible numbers only.

>
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number  :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧

>
Not correct if
there are all natural numbers such that
no further one exists below ω.

>
You seem to be saying:
| Not correct if
| not all natural numbers have
| a further one below ω

>
All further numbers are multiplied too.

ω is the first.infinite.ordinal.
The numbers.before.ω are all and only
0 and
any number.after.0 k such that
k and all non.0 numbers.before.k
have predecessors.
No number k⁺¹ exists such that
k and all non.0 numbers.before.k
have predecessors
but NOT
k⁺¹ and all non.0 numbers.before.k⁺¹
have predecessors
Thus,
no number.before.ω k exists such that
k⁺¹ is NOT a number.before.ω.
No numbers.before.ω k,m⁺¹ exists such that,
k+m is a number.before.ω
but NOT
k+m⁺¹ = (k+m)⁺¹ is a number.before.ω.
Thus,
no numbers.before.ω k,m exist such that
k+m is NOT a number.before.ω.
No numbers.before.ω k,m⁺¹ exists such that,
k⋅m is a number.before.ω
but NOT
k⋅m⁺¹ = (k⋅m)+k is a number before ω
Thus,
no numbers.before.ω k,m exist such that
k⋅m is NOT a number.before.ω.

In other words [1]:
| Not correct if
| ω is finiteⁿᵒᵗᐧᵂᴹ.

>
Not correct if
there is no free space between ℕ and ω.

ℕ = ⟦0,ω⦆
ω is after Avogadroᴬᵛᵒᵍᵃᵈʳᵒ
Avogadroᴬᵛᵒᵍᵃᵈʳᵒ = 6.02214076E23⁶ᐧ⁰²²¹⁴⁰⁷⁶ᴱ²³
and after any other number such that
it and all non.0 numbers.before.it
have predecessors.
Infiniteⁿᵒᵗᐧᵂᴹ is different.