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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: How many different unit fractions are lessorequal than all unit
 fractions?
Date: Wed, 9 Oct 2024 12:38:14 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:
> On 08.10.2024 19:28, Jim Burns wrote:
>> On 10/8/2024 6:18 AM, WM wrote:
> 
>>> All infinite endsegments contain more than any finite set of numbers.
>> ...still true if 'infinite' means "very large".
> No. Very large is not more than any finite set.

>> 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).

>> 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.

> Proof: If not, then there would be at least one endsegment with less
> numbers.
No. Why do you think that?

-- 
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.