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On 1/8/25 9:31 AM, WM wrote:On 08.01.2025 14:31, Jim Burns wrote:
>⦃k: k < ω ≤ k+1⦄ = ⦃⦄Let us accept this result.
ω-1 does not exist.
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Then the sequence of endsegments loses every natnumber but not a last one. Then the empty intersection of infinite but inclusion monotonic endsegments is violating basic logic. (Losing all numbers but keeping infinitely many can only be possible if new numbers are acquired.) Then the only possible way to satisfy logic is the non-existence of ω and of endsegments as complete sets.
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It is useless to prove your claim as long as you cannot solve this problem.
There isn't a "last one" to loseIf all are to lose and never more than one, then there is a last one to lose.
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