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

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.

 Each natural number k
is followed by successor k+1
k is not.in infinite E(k+1)

>
But infinitely many natnumbers are in e very infinite endsegment.

 There are no other end segments, none are finite.

>
They all have an infinite intersection.
>

The intersection of all end.segments is
the set of natural numbers in each end.segment.
Each natural number is not.in one or more end.segment.

>
That is true but wrong
in case of infinite endsegments which are infinite
because they have not lost all natural numbers.

 The intersection of all end.segments is
the set of natural numbers in each end.segment.

>
Infinitely many are in every infinite endsegment.

 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'.

>
Fine. All that blather does not contradict the fact that infinite endsegments are infinite.
>
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.