Sujet : Re: Replacement of Cardinality
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 19. Aug 2024, 20:06:44
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <ab0252f1-21b0-4c77-917c-37d0525baabb@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 8/19/2024 7:44 AM, WM wrote:
Le 17/08/2024 à 23:44, Jim Burns a écrit :
On 8/17/2024 9:36 AM, WM wrote:
NUF(x) at no point grows by more than 1.
>
At unit.fractions. Not at 0.
>
But NUF(x) grows.
>
At 0. Not at unit fractions.
>
You are not entitled to change the definition of NUF(x).
NUF(x) ≥ NUFᵈᵉᶠ(x) = |⅟ℕᵈᵉᶠ∩(∞,x]|
NUF(0) = NUFᵈᵉᶠ(0) = 0
∀ᴿx > 0: NUF(x) ≥ NUFᵈᵉᶠ(x) > 0
NUF(x) grows at 0
β = glb.{ y∈ℝ: ⅟ℕᵈᵉᶠ∩(∞,y]≠{} }
β = 0
⎛ Assume β > 0
⎜ not.bound 2⋅β > ⅟k ∈ ⅟ℕᵈᵉᶠ
⎜ not.bound ½⋅β > ¼⋅⅟k ∈ ⅟ℕᵈᵉᶠ
⎜ bound ½⋅β < β
⎝ Contradiction.
Therefore, β = 0
∀ᴿx > 0: NUF(x) ≥ NUFᵈᵉᶠ(x) > 0
NUF(x) grows at 0
NUF(x) grows at every unit fractions by 1.
No. ℵ₀+1 = ℵ₀
At 0 there is no unit fractions,
therefore NUF(0) = 0
as for all negative arguments.
The firstincrease happens at the first unit fraction.
No positive lower.bound, darkᵂᴹ or visibleᵂᴹ, exists of
visibleᵂᴹ unit.fractions.
Re: Replacement of Cardinality
Re: Replacement of Cardinality
