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

On 10/8/2024 6:18 AM, WM wrote:

On 07.10.2024 20:05, Jim Burns wrote:

That creates a problem (but for only you).
Your attempt to fix the problem involves
a change of the definition of intersection, or
of natural number, or of something else.

>
Not at all.

A natural number k can be counted.to from 0
For each split of naturals from 0 to k
the foresplit ends at some j and
the hindsplit begins at j+1
Each set S of natural numbers
either holds a first number min.S or is empty.
Each natural number k
has a predecessor.natural k-1 or is 0
Each natural number k
has a successor.natural k+1
You say I'm wrong about that.
That means I'm wrong about _what you mean_
But, if I'm wrong about what you mean,
then you're wrong about _what I mean_
and I'm just reporting common knowledge,
like the day of the week.

Changing what you think "infinite" means
from what we mean by "very large finite"
to what we mean by "infinite"

>
No.
Infinite means infinite.

What we mean by 'infinite'
is not
what you mean by 'infinite.
You use the same glossary, but
you use a different dictionary (if any).

All infinite endsegments cotain
more than any finite set of numbers.

...still true if 'infinite' means "very large".
There is more to being finite than that.
A finite set A can be ordered so that
each subset S
either holds first min.S and last max.S
or is empty.
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.
Each natural number k
is followed by successor k+1
k is not.in infinite E(k+1)
There are no other end segments, none are finite.

My argument requires only one case,
best demonstrated with endsegments E(n).
The intersection of all *infinite* endsegments
is infinite, because
they all contain the same natural numbers which
have not yet been eliminated by
the process E(n+1) = E(n) \ {n}.

>
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.
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?

Why else should they be infinite?

Because each natural number is followed by
a natural number,
because 'infinite' DOES NOT mean 'very large'.