Re: how

Le 03/06/2024 à 20:34, Jim Burns a écrit :

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⁺

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

Hence assumption (2) contradicts logic. Only one of the two complementary sets can and must contain the first point.

 

 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

No logic is admitted.

 

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.

It contains no logic.

 

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.

Not in a mathematics based upon logic.
Regards, WM