Re: Replacement of Cardinality

On 8/25/2024 3:45 PM, WM wrote:

Le 24/08/2024 à 19:06, Richard Damon a écrit :

On 8/23/24 10:49 AM, WM wrote:

NUF(x) increases from 0 to more.
But it  cannot increase by more than 1
without shzowing a constant level.

>
And,
since there is no finite value of x that
makes NUF(x) equal to 1, since for every finite x, their exists an x/2

>
For all visible x but not for all dark x.
>

Just because you have verbally defined a function
doesn't mean it exists and has proper values.

>
This function exists because
nothing contradicts its existence.

Except for the contradicting consequences of
its existence.
∃ᴿx>0: ¬∃¹ᐟᴺu<x  ⟹
∃ᴿβ≥x>0:
(∀ᴿγ>β: ∃¹ᐟᴺu<γ>β  ∧  ∀ᴿα<β: ¬∃¹ᐟᴺu<α<β)  ∧
(∃¹ᐟᴺu<½⋅β: 4⋅u<2⋅β>β  ∧  ¬∃¹ᐟᴺu<½⋅β<β) [contradiction]
Therefore,
¬∃ᴿx>0: ¬∃¹ᐟᴺu<x
¬∃ᴿx>0: NUF(x) = 1