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On 09.10.2024 12:12, FromTheRafters wrote:You really mean as long as only finitely many numbers are missing.WM presented the following explanation :Inclusion-monotony proves that all infinite endsegments have a commonTheorem: If every endsegment has infinitely many numbers, thenA conjecture is not a proof. This one is simply another non sequitur.
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.
infinite subset because only a loss of elements is possible. As long as
all endsegments are infinite, the loss has spared an infinite set common
to all.
If you can't understand try to find a counterexample.Notice anything? You should write it like this:
Or use a finite example.
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