Re: Replacement of Cardinality

Am 30.08.2024 um 09:21 schrieb joes:

Am Thu, 29 Aug 2024 13:37:03 +0000 schrieb WM:

Le 28/08/2024 à 17:06, joes a écrit :

Am Wed, 28 Aug 2024 13:04:57 +0000 schrieb WM:

Le 28/08/2024 à 06:20, joes a écrit :

Am Tue, 27 Aug 2024 19:57:39 +0000 schrieb WM:

>

The interval (0, x) cannot contain ℵo unit fractions unless x is
sufficiently large.

„Sufficient” meaning „not infinitesimal”.

No. "Sufficient" means that ℵo non-empty finite gaps between unit
fractions fit into the interval. That is much larger than
infinitesimal.

So… you agree.

Yes, I misread.

Thus: The interval (0, x) contains finitely many unit fractions only
for infinitesimal x.

Actually, it either contains NO unit fractions (if x = 0 or x is infinitesimal but > 0) or it contains infinitely many (ℵo) unit fractions.
Proof: If x = 0 then no unit fraction is in (0, x) (since in this case (0, x) is empty). If x is infinitesimal but > 0, then (by definition) for all n e IN: x < 1/n. Hence no unit fraction is in (0, x). If on the other hand x e IR, x > 0, and there is an unit fraction u such that u < x, then (0, x) contains (at least) the infinitely many unit fractions u, 1/(1/u + 1), 1/(1/u + 2), 1/(1/u + 3), ... (hence it contains ℵo unit fractions). qed
So no matter if x > 0 is infinitesimal or not: NUF(x) =/= 1.
Hence especially for all x e IR, x > 0: NUF(x) =/= 1. In fact, for all x e IR: NUF(x) e {0, aleph_0}.
WM is a crank.