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

On 10/9/2024 4:11 AM, WM wrote:

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.

There are two dictionaries in use here,
_yours_ and _ours_
Yes,
_your_ very large
is not more than any
_your_ finite set.
Also,
_our_ very large
is not more than any
_our_ finite set.
That isn't the difference
between dictionaries.
----
Anything with slightly more than
_our_ finite is also
_our_ finite.
You want to qualify that with 'dark'
'Dark' is in _your_ dictionary.
_Ours_ doesn't distinguish 'dark' and 'visible'.
Dark.or.visible anything with slightly more than
dark.or.visible _our_ finite is also
dark.or.visible _our_ finite.
----
⎛ Anything with slightly more than
⎜ _our_ finite is also
⎝ _our_ finite.
Turn it around and get
⎛ Anything with slightly fewer than
⎜ _our_ infinite is also
⎝ _our_ infinite.
In _your_ 'mathematics and logic'.
which is _your_ dictionary,
infinite is countable.down.from
to finite through dark.
That's not _our_ dictionary.
In _our_ dictionary
infinite is NOT countable.down.from
to finite through dark.or.visible.
There isn't any finite
  slightly below infinite.
There isn't any dark
  slightly below infinite.
There isn't a gap
  slightly below infinite.
There isn't a there
  slightly below infinite.

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.
All.the.end.segments have an empty 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.

Zero.many are in
the intersection of all.the.end.segments,
each of which is infinite.

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.

⎛ By definition, dark or visible,
⎜ their sets are minimummed or empty, and
⎜ they are predecessored or 0, and
⎝ they are successored.
⎛ Each upper.bounded.set A≠{} of them, {k≥ᵉᵃᶜʰA}≠{}
⎜ has a minimum upper.bound min.{k≥ᵉᵃᶜʰA}
⎜
⎜ Either min.{k≥ᵉᵃᶜʰA} is in A≠{}
⎜ or (min.{k≥ᵉᵃᶜʰA})-1 is
⎜ a smaller.than.minimum upper.bound: gibberish.
⎜
⎜ min.{k≥ᵉᵃᶜʰA} is in A≠{}
⎜ min.{k≥ᵉᵃᶜʰA} = max.A exists
⎜ min.A≠{} exists.
⎜
⎜ Upper.bounded A≠{} is two.ended.
⎜ By a similar argument,
⎜ each subset S≠{} is two.ended.
⎜
⎝ Upper.bounded A≠{} is finite.
⎛ Not.upper.bounded A≠{} is not.two.ended.
⎝ Not.upper.bounded A≠{} is not.finite.
Set A≠{} of natural numbers
is finite iff it is upper.bounded.

Why else should they be infinite?

>
Because each natural number is followed by
a natural number,

and thus is not.upper.bounded and not.finite.

not only one but infinitely many

Right.
Because each natural number is followed by
infinitely.many minima of
infinitely.many end.segments which it isn't in.
But one which it isn't in is enough to be
not.in the intersection of all.

because 'infinite' DOES NOT mean 'very large'.

>
Fine.

And no 'slightly.fewer.than.infinite' exists,
neither in the dark nor in the visible.

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.

Can you show that without a quantifier shift,
which is unreliable?