Re: because g⤨(g⁻¹(x)) = g(y) [1/2] Re: how

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Sujet : Re: because g⤨(g⁻¹(x)) = g(y) [1/2] Re: how
De : noreply (at) *nospam* example.com (joes)
Groupes : sci.math
Date : 03. May 2024, 20:27:18
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v13dqm$2uk84$5@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Fri, 03 May 2024 13:29:46 +0000 schrieb WM:

Le 01/05/2024 à 23:46, Moebius a écrit :
Am 30.04.2024 um 15:12 schrieb WM:
If n is before ω then n⋅2 is before ω. (*)
 
That is not true.
 
Wir haben mithin also gezeigt, dass "If n is
before ω then n⋅2 is before ω" gilt. qed
 
Your proof holds for all natnumbers that can be found - a tiny initial
segment of ℕ.
 
Every findable natnumber has ℵo natsuccessors. ==> Not all can be found.
All natural numbers have no natsuccessors. ==> There are more natnumbers
than can be found.
But they can only be applied collectively.

The fuck are „findable” naturals?
Every natural number has a (also natural) successor.

--
joes

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