Re: how

On 5/27/2024 10:03 AM, WM wrote:

Le 26/05/2024 à 23:21, Jim Burns a écrit :

On 5/26/2024 3:15 PM, WM wrote:

Le 25/05/2024 à 19:23, Jim Burns a écrit :

On 5/23/2024 8:10 AM, WM wrote:

There is no unit fraction smaller than all x > 0,
[A]

>
Also true:
There is no x > 0 smaller than all unit fractions.
[B]

>

Note that
points on the real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.

ℝ is the set holding
points.between.sides of non.∅.splits of ℚ
ℚ is the set holding
differences.of.quotients of ℕ
ℕ is the set holding
the decrementable ordinals and 0
which are preceded by only
decrementable ordinals and 0
Note that
points on the positive real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.

Agreed.
>

If A is true, then there is
a positive non-unit fraction smaller than
all unit fractions.

>
No.
>
| Assume that claim is correct.
| Assume that x > 0 is smaller than
| all unit.fractions.
|
| However,
| ⅟⌊(1+⅟x)⌋ is a unit.fraction < x
| Contradiction.

>
Not possible with dark number x.

x ∈ ℝ
x is between foresplit hindsplit Fₓ Hₓ of ℚ
0 < x
0 ∈ Fₓ ⊆ ℚ
If 0 were max Fₓ
then
0 would be between Fₓ Hₓ
and 0 = x
But 0 < x
so 0 is NOT max in Fₓ ⊆ ℚ
Fₓ\(∞,0] is not empty
rₓ/sₓ ∈ Fₓ\(∞,0]
0 < rₓ/sₓ < x  and  rₓ/sₓ ∈ ℚ
⅟⌊(1+sₓ/rₓ)⌋ is a unit.fraction < x

Not possible with dark number x.

Then darkᵂᴹ numbers aren't in ℝ

That claim is incorrect.

>
for visible numbers.

Incorrect for ℝ
(points.between.sides of non.∅.splits of ℚ)
Therefore darkᵂᴹ numbers aren't in ℝ

If B is true, then there is
a unit fraction smaller than
all positive non-unit fraction.

>
No.
>
| Assume that claim is correct.
| Assume that ⅟n is smaller than
| all positive non.unit.fractions.
|
| However,
| ⅟(n+π) is a non.unit.fraction < ⅟n
| Contradiction.

>
Not possible with dark number n.

n ∈ ℕ ⊆ ℚ ⊆ ℝ
n+π ∈ ℝ\ℚ
n < n+π
n⋅⅟n < (n+π)⋅⅟n
⅟(n+π)⋅1 < ⅟(n+π)⋅(n+π)⋅⅟n
⅟(n+π) < ⅟n
⅟(n+π) ∈ ℝ\ℚ
⅟(n+π) ∉ ⅟ℕ ⊆ ℚ

That claim is incorrect.

>
for visible numbers.

Then darkᵂᴹ numbers aren't in ℕ

There is no third alternative.

>
Consider ⅟⌊(1+⅟x)⌋ and ⅟(n+π)

>
I consider that
points on the real axis are fixed and
can be subdivided into two sets.

ℝ is the set holding
points.between.sides of non.∅.splits of ℚ
ℚ is the set holding
differences.of.quotients of ℕ
ℕ is the set holding
the decrementable ordinals and 0
which are preceded by only
decrementable ordinals and 0
Note that
points on the positive real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.

The decision is that not all subsets
of unit.fractions or of non.unit.fractions
have two ends.

>
The real axis and all point sets in it
have an end at or before zero.

The positive real axis ℝ⁺
does not hold a visibleᵂᴹ lower end.
0 ∉ ℝ⁺
¬∃y ∈ ℝ⁺: ¬∃x ∈ ℝ⁺: y > x = y/2
Not all non.∅.subsets of visibleᵂᴹ ℝ⁺ hold two ends.
ℝ⁺ doesn't.
For each darkᵂᴹ.number.holding superset ℝ⁺∪𝔻ᵂᴹ
not all non.∅.subsets of ℝ⁺∪𝔻ᵂᴹ hold two ends.
ℝ⁺ doesn't.
Holding two ends is insufficient for
a set to be finite.
Each non.∅.subset holding two ends is
necessary and sufficient for a set to be finite.