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Subject: Re: how
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Date: Wed, 22 May 24 17:57:28 +0000
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From: WM <wolfgang.mueckenheim@tha.de>
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Le 22/05/2024 à 17:48, Jim Burns a écrit :
> On 5/22/2024 6:43 AM, WM wrote:
>> Le 22/05/2024 à 01:27, Jim Burns a écrit :
> 
>>> ℝ is ℚ and points between non.∅ splits of ℚ 
> 
>>> for any x > 0
>>
>> that you can determine
> 
> For any x > 0 in ℚ or between a non.∅ split of ℚ
> more.than.any.k<ℵ₀ unit.fractions
> sit before x
> among them are ⅟⌊(1+sₓ/rₓ)⌋ to ⅟⌊(k+1+sₓ/rₓ)⌋
> 0 < rₓ/sₓ < x

Between each pair of unit fractions, there is a finite distance (with many 
points x > 0).
Hence not even two unit fractions can satisfy the condition to sit before 
any x > 0.
Therefore Ax > 0: NUF(x) = ℵo is wrong.
> 
>> If
>> there is no unit fraction smaller than all x > 0,
>> then
>> there is an x > 0 preventing this.
>
> There is no x > 0 smaller than all unit fractions.
> ¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k

There is an x >= 0 smaller than all unit fractions.
> 
> | Assume otherwise.
> | Assume x¹ᐟᴺ > 0:  ∀¹ᐟᴺ ⅟k: x¹ᐟᴺ ≤ ⅟k

That does not destroy this condition:
Between each pair of unit fractions,
there is a finite distance.
Hence not even two unit fractions can satisfy the condition to sit before 
any x > 0.
Therefore Ax > 0: NUF(x) = ℵo is wrong.

Why do you never address this fact?

Of course if all numbers were visisble, we had a contradiction. That does 
not change by your repeated proofs of this fact. 

Try to refute this fact: Between each pair of unit fractions, there is a 
finite distance. Therefore of many unit fractions, all but at most one are 
not smaller than every x > 0. Hence
Ax > 0: NUF(x) = ℵo is wrong.

Regards, WM