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From: WM <wolfgang.mueckenheim@tha.de>
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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