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

On 09.10.2024 14:38, joes wrote:

Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:

An end segment E is infinite because each natural number has a
successor,
so no element of E is max.E so not all nonempty subsets are two.ended
so E is not finite.

Therefore every infinite endsegment has infinitely many elements with
each predecessor in common. This is valid for all infinite endsegments.

With each, but not with all at once (cf. quantifier shift).

With all infinite endsegments at once! Inclusion monotony. If you can't understand try to find a counterexample.

 

There are no other end segments, none are finite.

They all have an infinite intersection.

Intersection with what?
 

Consider end segment E(k+1)
k+1 is in E(k+1)
Is k+1 in each end segment? Is k+1 in E(k+2)?
Is E(k+1) the set of natural numbers in each end.segment?

No, but by definition there are infinitely many numbers. They are dark.

 

Why else should they be infinite?

Because each natural number is followed by a natural number,

not only one but infinitely many

 

because 'infinite' DOES NOT mean 'very large'.

Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.

Invalid quantifier shift.

Valid quantifier shift.

 

Proof: If not, then there would be at least one endsegment with less
numbers.

No. Why do you think that?

The shrinking endsegments have all their elements in common with all their predecessors. As long as all are infinite, then all have an infinite set in common.
Regards, WM