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

On 10.10.2024 21:51, joes wrote:

Am Wed, 09 Oct 2024 15:24:21 +0200 schrieb WM:

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.

Fail.

There are no other end segments, none are finite.

They all have an infinite intersection.

Intersection with what?

Only the intersection of finitely many segments is infinite.

Of course. If infinitely many numbers are indices, then infinitely many cannot be contents.

 

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.

You haven't proved your logical rule in general. In particular, you can't
use it here to argue it is valid, that would be circular.

The proof of the cake lies in the eating.
I do not intend to prove quantifier shift in general. I prove: If every endsegment is infinite, then infinitely many numbers are in all endsegments of this subset.

Can you explain what a quantifier shift is?

The above arguing. If every endsegment has an infinite subset, then there exists one and the same infinite subset of every endsegment. Proof: Inclusion monotony.
Regards, WM