Re: How many different unit fractions are lessorequal than all unit fractions?

On 09.10.2024 16:40, FromTheRafters wrote:

WM pretended :

On 09.10.2024 12:12, FromTheRafters wrote:

WM presented the following explanation :

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Theorem: If every endsegment has infinitely many numbers, then infinitely many numbers are in all endsegments.
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Proof: If not, then there would be at least one endsegment with less numbers.

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A conjecture is not a proof. This one is simply another non sequitur.

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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.

Have you succeeded yet?

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}
...

 Finite sets again, I don't care how many.
 

As long as five numbers remain in all sets, they are a common subset of all sets.
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When you add all numbers following 10 to all sets, the situation remains the same.

 Try using infinite sets.

The completed sets are infinite.

If there were a last as well as a first you would be right, but there is no last to be in all endsegments,

The same numbers beyond any n are in all infinite endsegments. There are infinitely many because there is no last number.
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so it is empty.

Infinitely many numbers in all infinite endsegments do not make an empty intersection. An infinite set can follow only on a finite number n.
Regards, WM