Re: how

Liste des GroupesRevenir à s math 
Sujet : Re: how
De : invalid (at) *nospam* example.invalid (Moebius)
Groupes : sci.math
Date : 13. Jun 2024, 16:23:07
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v4f2sr$2aak9$5@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11
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Am 13.06.2024 um 17:07 schrieb WM:
Le 13/06/2024 à 16:53, Moebius a écrit :
Am 13.06.2024 um 16:39 schrieb WM:
Le 13/06/2024 à 14:50, Moebius a écrit :
Am 13.06.2024 um 14:22 schrieb Jim Burns:
On 6/12/2024 4:33 PM, WM wrote:
>
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.
>
Observation: If "definable numbers" is replaced by "finitely many numbers"
>
By definition definable implies existence of a FISON.
>
Ich glaube, Du verwechselst hier gerade "definable" mit "defined".
>
Aber natürlich impliziert "defined" "definable". :-)
>
(Denn wenn eine Zahl nicht "definable" wäre, könnte sie auch nicht "defined" sein.)
>
Hinweis: Da die natürlichen Zahlen nach von Neumann FISONs SIND, sind diese also auch "definable" bzw. "defined".
 Of course.
>
Damit ist dann aber klar, dass die Behauptung "If only definable numbers are subtracted, then successors remain." falsch ist.
 No.
Yes. (Hinweis: Du bist offensichtlich für jede Art von Mathematik zu dumm, Mücke.)
Hier zur Erklärung: Wenn IN = IN_definable , dann ist IN \ IN_definable = IN \ IN = { }.
Mit anderen Worten: Die Behauptung "If only definable numbers are subtracted, then successors remain" ist falsch, da in { } offensichtlich _nichts_ "remains", also insbesondere auch keine "successors".

"Richtig" dagegen wäre die Behauptung: "If only finitely many numbers are subtracted, then successors remain."
 All FISONs or v. Neumann ordinals [in IN] are finite.
Äh ja, das ist allgemein bekannt. Hint: The "F" in "FISON" means /finite/.

Theorem: More than finitely many finite initial segments cannot be merged.
Aha: "merged". :-)

Proof: This is caused by the pigeonhole principle and the definition "finite initial segment". If each of the first n positive integers has a unary representation in form of a string, like ooooo, that is shorter than n then, by the pigeonhole principle, there must be two different positive integers defined by the same unary representation.
Huh?!

Clearly this is absurd.
Agree! This "proof" is nonsense.

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