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On 15.02.2025 18:38, joes wrote:Am Sat, 15 Feb 2025 15:55:55 +0100 schrieb WM:On 15.02.2025 14:40, Richard Damon wrote:
No. It proves that it doesn’t matter *which single element* is removed.Induction proves an inductive set F of FISONs that can be removedIndeed, if you remove any number of FISONs, infinitely many remain.The set you show is empty, is the set of FISONs that are individuallyThe remaining set should have at least one element.
not necessary.
without changing the result U(F) = ℕ ==> U(F\F) = ℕ.
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