Re: Replacement of Cardinality

On 8/29/24 9:26 AM, WM wrote:

Le 28/08/2024 à 22:35, Moebius a écrit :
 

Hint: NUF(x) := card({s e SB : s < x})   (x e IR)
>
Hence NUF(x) = 0 for all x e IR, x <= 0, and NUF(x) = aleph_0

  for all x > the minimum distance between many unit fractions which is not 0.
 Regards, QM
 

No, for all x > 0, since there is no finite minimum distance between any two unit fractions, as for any attempt to come up with one, there will be two unit fractions closer than that.
Note, we aren't talking about "many" unit fractions, but ALL unit fractions.
Your "many" is just a dodge to get around the fact that your logic just can't handle them all.