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

Am Wed, 09 Oct 2024 16:50:55 +0200 schrieb WM:

On 09.10.2024 16:40, FromTheRafters wrote:

WM pretended :

On 09.10.2024 12:12, FromTheRafters wrote:

WM presented the following explanation :

>

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

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

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.

Only as long as. Then we just get some E(n) for some finite n.

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.

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

The completed sets are infinite.

Can you write that out for me?

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.

Exactly. There is no last n.

so it is empty.

Infinitely many numbers in all infinite endsegments do not make an empty
intersection.

NB there are infinitely many endsegments (for each element...).

An infinite set can follow only on a finite number n.

Never contradicted.
--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.