Re: Replacement of Cardinality

On 8/5/2024 3:04 PM, WM wrote:

Le 04/08/2024 à 22:16, Jim Burns a écrit :

On 8/4/2024 2:13 PM, WM wrote:

More of interest are these two claims which are
not both true or both false:
For every x there is u < x.
There is u < x for every x.
The latter is close to my function:
There are NUF(x) u < x.

>

 From ∀x∃U to ∃U∀x is unreliable,

>
There is no from to.
NUF(x) is so defined.

NUFᵉᵃᶜʰ  :=  |{u∈⅟ℕᵈᵉᶠ: u <ᵉᵃᶜʰ ℝ⁺}|
NUFᵈᵉᶠ(x)  :=  |{u∈⅟ℕᵈᵉᶠ: u < x}|
∀v ∈ ⅟ℕᵈᵉᶠ:
⎛ ¬(v < u/2 ∈ ℝ⁺)
⎜ ¬(v <ᵉᵃᶜʰ ℝ⁺)
⎝ ¬(v ∈ {u∈⅟ℕᵈᵉᶠ: u <ᵉᵃᶜʰ ℝ⁺})
NUFᵉᵃᶜʰ = |{}| = 0
⅟ℕᵈᵉᶠ∩(0,1] is maximummed.
⅟ℕᵈᵉᶠ∩(0,x) is maximummed.
⅟ℕᵈᵉᶠ∩(0,1] is down.stepped.
⅟ℕᵈᵉᶠ∩(0,x) is down.stepped.
⅟ℕᵈᵉᶠ∩(0,1] is non.max.up.stepped.
⅟ℕᵈᵉᶠ∩(0,x) is non.max.up.stepped.
⅟ℕᵈᵉᶠ∩(0,1] is ℵ₀.many.
⅟ℕᵈᵉᶠ∩(0,x) is ℵ₀.many.
NUF(1) = ℵ₀
NUF(x) = ℵ₀

For every number of unit fractions
NUF(x) gives the smallest interval (0, x).

For  each real x > 0
for each finite cardinal k > 0
there is a finite.unit.fraction uₖ such that
|⅟ℕᵈᵉᶠ∩[uₖ,x)| = k
uₖ = ⅟⌊k+⅟x⌋
For  each x > 0
for each finite cardinal k > 0
there is a finite.unit.fraction uₖ such that
|⅟ℕᵈᵉᶠ∩[uₖ,x)| = k
⅟(1+⅟uₖ) ∈ ⅟ℕᵈᵉᶠ∩(0,uₖ)
k < |⅟ℕᵈᵉᶠ∩(0,x)|
For  each x > 0
for each finite number k
k ≠ |⅟ℕᵈᵉᶠ∩(0,x)|
For  each x > 0
⅟ℕᵈᵉᶠ∩(0,x) is not finite.