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

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.

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.

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.
Can you explain what a quantifier shift is?

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.

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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.