Re: Replacement of Cardinality

On 8/3/2024 7:25 AM, WM wrote:

Le 02/08/2024 à 19:31, Moebius a écrit :

For each and every of these points [here referred to with the variable "x"]: NUF(x) = ℵ₀ .

 I recognized lately that you use the wrong definition of NUF.
Here is the correct definition:
There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
Note that the order is ∃ u ∀ y.
NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already is wrong since there is no unit fraction smaller than all unit fractions.
ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

>
0->(...)->(1/1)
>
Contains infinite unit fractions.
>
0->(...)->(1/2)->(1/1)
>
Contains infinite unit fractions.
>
0->(...)->(1/3)->(1/2)->(1/1)
>
Contains infinite unit fractions.
>
However, (1/3)->(1/1) is finite and only has three unit fractions expanded to:
>
(1/3)->(1/2)->(1/1)
>
Just like the following has four of them:
>
(1/4)->(1/3)->(1/2)->(1/1)
>
>
(0/1) is not a unit fraction. There is no smallest unit fraction. However, the is a largest one at 1/1.
>
A interesting part that breaks the ordering is say well:
>
(1/4)->(1/2)
>
has two unit fractions. Then we can make it more fine grain:
>
(1/4)->(1/2) = ((1/8)+(1/8))->(1/4+1/4)
>
;^)