Re: There is a first/smallest integer (in Mückenland)

Am Sat, 20 Jul 2024 12:27:11 +0000 schrieb WM:

Le 19/07/2024 à 16:29, Moebius a écrit :

Am 19.07.2024 um 15:53 schrieb WM:

Le 18/07/2024 à 23:35, Moebius a écrit :

Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :
>

         NUF(x) = aleph_0   for all x e IR, x > 0 , (*)

That means

that for each and every x e IR, x > 0 there are aleph_0 unit
fractions which are <= x.

Then we can abbreviate each and every x > 0 by (0, oo).

Huh?!

If something is left of all x > 0 then it is left of the interval (0,
oo)

Nobody is claiming that.

Ich glaube, das hat man Dir jetzt schon so um die 500- bis 1000-mal
erklärt,

No explanations of your quantifier nonsensebut a counter example please.

(0, oo) contains nothing but all x > 0.

Ja, mückenheim, das Intervall (die Menge) (0, oo) enthällt genau alle
reellen Zahlen > 0.
In Zeichen: x e (0, oo) <-> x e IR & x > 0.
Das bedeutet (ins Unreine gesprochen), dass man in Formeln/Aussagen den
Ausdruck "x e IR & x > 0" durch den Ausdruck "x e (0, oo)" ersetzen
kann und umgekehrt.

One of these is the result left-hand side of every x > 0 <==> left-hand
side of (0, oo).

It is not the same y that is smaller than every x.

Man kann also z. B. (*) auch so schreiben:
             NUF(x) = aleph_0   for all x e (0, oo) .
 
Rein formal solle man aber die Formeln SO schreiben:
             for all x e IR, x > 0: NUF(x) = aleph_0 ,
bzw.
             for all x e (0, oo): NUF(x) = aleph_0 .

 
That means NUF(x) is constant over the whole interval (0, oo), i.e., a
false expression because it follows that NUF increases from 0 to
infinity between [0, 1] and (0, 1].

No rule saying a function has to be continuous.

--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.