Re: Replacement of Cardinality

On 8/25/24 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 visisble x but not for all dark x.

But all finite values ARE visible, as for ANY finite x, there exists an x/2.
Your "Dark" numbers are just not finite values.

 

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.
 Regards, WM

Except that there is not value for it to be 1 at.
You can't use the assumption of the existance of the function to prove that it must exist, especially when that creates the contradictions.
All you are proving is that you logic can't have the Natural Numbers as part of there domain, and thus ANYTHING you try to prove with your logic must be in a set smaller than that, and to claim otherwise is just a LIE.