Re: Replacement of Cardinality

On 8/30/24 9:06 AM, WM wrote:

Le 30/08/2024 à 03:09, Richard Damon a écrit :

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

Le 28/08/2024 à 22:35, Moebius a écrit :
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Hint: NUF(x) := card({s e SB : s < x})   (x e IR)
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Hence NUF(x) = 0 for all x e IR, x <= 0, and NUF(x) = aleph_0

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for all x > the minimum distance between many unit fractions which is not 0.
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No, for all x > 0,

 Easier to answer: Are there two unit fractions lesorequal than all unit fractions?
 Regards, WM
 

No, but there are two unit fractions less than any given unit fraction, which is all that is needed to show that NUF(x) is broken, and can't have the value 1 at any finite unit fraction.