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On 09.10.2024 12:12, FromTheRafters wrote:Finite sets again, I don't care how many.WM presented the following explanation :>>Theorem: If every endsegment has infinitely many numbers, then infinitely many numbers are in all endsegments.A conjecture is not a proof. This one is simply another non sequitur.
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Proof: If not, then there would be at least one endsegment with less numbers.
Inclusion-monotony proves that all infinite endsegments have a common 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.
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If you can't understand try to find a counterexample.
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Or use a finite example.
Diminish the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} to get
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
...
As long as five numbers remain in all sets, they are a common subset of all sets.Try using infinite sets. If there were a last as well as a first you would be right, but there is no last to be in all endsegments, so it is empty.
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When you add all numbers following 10 to all sets, the situation remains the same.
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