Re: how

On 6/3/2024 7:58 AM, WM wrote:

Le 03/06/2024 à 10:57, Jim Burns a écrit :

⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y   implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠xv : x<y

>
No this is not implied but
independently proven in Evidence for Dark Numbers,
prepublished chapter 4.2:
>
We assume that
all 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.

2.
Or we can assume instead that
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
ℚ⁺ holds all.and.only ratios of numbers in ℕ⁺
ℝ⁺ holds all of ℚ⁺ and all.and.only
points x between open.foresplits Fₓ and ℚ⁺\Fₓ of ℚ⁺
with no points zero distance apart
and
⅟ℕ holds all.and.only reciprocals of numbers in ℕ⁺
ℝ⁺\⅟ℕ holds all.and only the others in ℝ⁺
and
ℝ⁺ is the positive real axis.

For visible numbers we have two statements both of which are true:
[A]
There is no unit fraction smaller than
all positive non-unit fractions.
[B]
There is no positive non-unit fraction smaller than
all unit fractions

Under assumption (2.)
[A] and [B] are provable for all of ⅟ℕ and ℝ⁺\⅟ℕ

 If A is true for dark numbers too,
then there is a positive non-unit fraction
smaller than all unit fractions.

Equivalent to:
If there is no positive non.unit.fraction
smaller than all unit fraction,
then A is false for darkᵂᴹ numbers too
And then, by (2.), no darkᵂᴹ numbers are in ℝ⁺\⅟ℕ

 If B is true for dark numbers too,
then there is a unit fraction
smaller than all positive non-unit fractions.

Equivalent to:
If there is no unit fraction
smaller than all positive non.unit.fractions,
then B is false for darkᵂᴹ numbers too
And then, by (2.), no darkᵂᴹ numbers are in ⅟ℕ

There is only one objection:

Versions of ℕ⁺ ℚ⁺ and ℝ⁺ which hold darkᵂᴹ numbers
are provably not the (2.) version.
Whatever is proved or claimed or hallucinated about
some other version is not a claim about
the (2.) version.

Not all subsets of unit fractions or
of non-unit fractions have two ends.

Pick a non.two.ended subset. 'Bye, Bob.

But this is dismissed by the fact that the positive real axis and all point sets in it
have an end at or before zero.

You're too late. Bob's gone.