Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

Am Mon, 28 Oct 2024 11:03:22 +0100 schrieb WM:

On 27.10.2024 19:04, joes wrote:

Am Sun, 27 Oct 2024 17:12:23 +0100 schrieb WM:

Am 27.10.2024 um 14:54 schrieb Moebius:

Am 27.10.2024 um 08:38 schrieb WM:
>

and after NUF(x') = 1

There is no x' e IR such that NUF(x') = 1.
Hint: For each and every x e IR, x <= 0: NUF(x) = 0 and for each and
every x e IR, x > 0: NUF(x) = aleph_0.

That is blatantly wrong because it would require that ℵo unit
fractions exist between 0 and each and every x > 0,

Which is obviously the case.

Between two unit fractions there are uncountably many x > 0. Therefore
your claim is wrong.

I can't see how that follows?

If there were a real x with finitely many UFs less than it, the
finitely many larger UFs... couldn't have infinitely many lesser UFs.
Unless you claim finitely many UFs.

There is a smallest UF.

No, there are inf. many.

i.e., the open interval (0, 1].

No, you shifted the quantifiers again.

This is the definition of NUF(x): There are NUF(x) UFs between 0 and x.
Not: For given x, there are infinitely many UFs.

That IS the case.

There is no UF
smaller than all x > 0 because every UF is an x > 0 itself.

Correct, there is no smallest UF >0. But the quantifier-shifted claim
was: there is a different UF y smaller than any single other UF x.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.