Liste des Groupes | Revenir à s math |
Am Wed, 26 Feb 2025 13:49:49 +0100 schrieb WM:
Then U(F) must contain more than all n ∈ U(F). But it does not. And you cannot find a difference.The power of induction proves ∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵoIt does not prove |N \ U{F(n): n e U(F)}| = oo.
Then show what remains after removing all n ∈ U(F).Indeed, that you keep referring to removing |N| elements insteadThe *set* is not in the set.No, why should it? But the set is proven to exist by induction.
of n e N is a mistake on your part. The existence has never been
disputed.
Les messages affichés proviennent d'usenet.