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Am Sat, 12 Oct 2024 18:51:10 +0200 schrieb WM:But not an infinite endsegment.On 10.10.2024 21:51, joes wrote:Am Wed, 09 Oct 2024 15:24:21 +0200 schrieb WM:On 09.10.2024 14:38, joes wrote:Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:What? There's a segment for every number.Of course. If infinitely many numbers are indices, then infinitely manyOnly the intersection of finitely many segments is infinite.Intersection with what?There are no other end segments, none are finite.They all have an infinite intersection.
cannot be contents.
It is true because there is no counterexample.I do not intend to prove quantifier shift in general. I prove: If everyThen you need to explain why the shifted proposition should be true.
endsegment is infinite, then infinitely many numbers are in all
endsegments of this subset.
Why not?Not valid for infinite sets.Can you explain what a quantifier shift is?The above arguing. If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.
Proof: Inclusion monotony.
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