Re: Replacement of Cardinality

On 8/29/2024 9:38 AM, WM wrote:

Le 28/08/2024 à 19:15, Jim Burns a écrit :

On 8/27/2024 3:29 PM, WM wrote:

Le 25/08/2024 à 23:28, Richard Damon a écrit :

Your final answer is basically just
admitting that your logic can't supply
the needed properties of the Natural Numbers.

>
No logic can treat
the complete set of natural numbers
without dark numbers.

>
∀S ⊆ ℕ: S ≠ {} ⇔ ∃k ∈ S: k = min.S
>
∀k ∈ ℕ: k ≠ 0 ⇔ ∃j ∈ ℕ: j+1 = k
>
∀j ∈ ℕ: ∃k ∈ ℕ: j+1 = k

>
That treats the potentially infinite collection.

Call it ℕᴾᴵꟲ.
The name doesn't matter.
There is no natural number not.in ℕᴾᴵꟲ.
⎛ Assume otherwise.
⎜ Assume that darkᴹᵂ 𝔊 ≥ᵉᵃᶜʰ ℕᴾᴵꟲ exists.
⎜
⎜ Consider the corresponding unit fractions.
⎜ ⅟𝔊 is between 0 and each visibleᴹᵂ in ⅟ℕᴾᴵꟲ
⎜ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ ⅟𝔊 > 0
⎜
⎜ However,
⎜ 0 = greatest.lower.bound.⅟ℕᴾᴵꟲ
⎜ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ ⅟𝔊 > 0
⎜
⎜ 𝔊 ≱ᵉˣⁱˢᵗˢ ℕᴾᴵꟲ
⎝ Contradiction.
Therefore,
there is no natural number not.in ℕᴾᴵꟲ.
ℕᴾᴵꟲ is
the complete flying.rainbow.sparkle.pony
of natural numbers.
----
0 = greatest.lower.bound.⅟ℕᴾᴵꟲ
0 > α  ⇒  ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ α
γ > 0  ⇒  ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ γ
⎛ Assume otherwise.
⎜ Assume greatest.lower.bound.⅟ℕᴾᴵꟲ = β > 0
⎜
⎜ β > ½⋅β
⎜ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ ½⋅β
⎜
⎜ 2⋅β > β
⎜ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ 2⋅β
⎜ exists visibleᵂᴹ ⅟k ≱ 2⋅β
⎜ exists visibleᵂᴹ ¼⋅⅟k ≱ ½⋅β
⎜ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ ½⋅β
⎝ Contradiction.
Therefore,
there is no higher lower.bound than 0
Also,
0 is a lower.bound.
⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ 0
0 = greatest.lower.bound.⅟ℕᴾᴵꟲ