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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: How many different unit fractions are lessorequal than all unit
 fractions?
Date: Fri, 6 Sep 2024 22:01:12 -0400
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On 9/6/24 8:12 AM, WM wrote:
> On 05.09.2024 21:57, Jim Burns wrote:
>> On 9/5/2024 9:59 AM, WM wrote:
> 
>>> NUF(x) increases from 0 to more.
>>
>> ...at 0.
> 
> No. NUF(x) counts the unit fractions between 0 and x. Between 0 and 0 
> there is no unit fraction. NUF(0) = 0.

But it CAN'T for the values of x that are finite, as for ANY finite 
number, there is an infinite number of unit factions below it, so it 
can't ever have a finite value for any finite x.

So, either NUF(x) counts some sub-finite "unit fractions" (that are not 
the reciprocals of Natural Numbers) or it just doesn't exist.

> 
>>> It cannot increase to 2 or more
>>> before having accepted 1.
>>
>> NUFᵈᵉᶠ(x) cannot increase to 2
>> without having already been ≥ 2
> 
> Impossible. NUF(0) = 0. There is a first increase in linear order.

Which can't be the reciprical of a Natural Numbers (since there will 
always be a smaller number) so it must be counting some sub-finite 
numbers as unit fractions or it just doesn't exist.

>>> It cannot increase to ℵo without
>>> having accepted 1, 2, 3, ...
>>
>> x > 0  ⇒
>> 0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x
> 
> 0 is smaller than all that. Therefore there is no increase at 0.
> x is larger than all that. Therefore your x is not the least one posiible.

So it can't increment finitely at recipricals of Natural Numbers so 
either it is counting some sub-finite values too, or it just doesn't exist.

> 
> Regards, WM
>