Re: Replacement of Cardinality

On 8/30/24 11:29 AM, WM wrote:

Le 30/08/2024 à 15:23, Richard Damon a écrit :

On 8/30/24 8:59 AM, WM wrote:

Le 30/08/2024 à 12:13, Moebius a écrit :

Am 30.08.2024 um 09:21 schrieb joes:

 

So no matter if x > 0 is infinitesimal or not: NUF(x) =/= 1.

>
Are there two unit fractions possible lessorequal than all unit fractions?
>

No, because nothing is smaller than itself, and two unequal order things must have an order.

 Not both can be the smallest. Hence only one is the first.

But since there WILL be another smaller than it, it won't be the smallest, just the smallest of those mentioned so far, not of all.

>
But there ARE two unit fractions less than ANY given unit fraction.

 Of course. Any given unit fraction can be given and therefore is visible.

And thus, none can be the smallest, as for *ANY* given unit fraction, there is a smaller one.
If there is something smaller than you, you can not be the smallest.

>
You just apply the wrong qualifiers because you logic system is broken.

 No. I apply this way in oder to show the more intelligent readers among you, than there is NUF(x) = 1.
 

No, it shows that there CAN'T be an finite x where NUF(x) = 1, as there will always be at least two unit fractions below it.

Regards, WM