Re: how

On 5/21/2024 12:55 PM, WM wrote:

Le 21/05/2024 à 17:50, Jim Burns a écrit :

On 5/20/2024 4:08 PM, WM wrote:

Le 19/05/2024 à 22:28, Jim Burns a écrit :

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.

>
For any x > 0
more.than.2 unit.fractions
sit before x
among them are ⅟⌊(1+⅟x)⌋ ⅟⌊(2+⅟x)⌋ ⅟⌊(3+⅟x)⌋

>
Of course,
for any x that you can name,
there are ℵo smaller unit fractions.

ℝ is ℚ and points between non.∅ splits of ℚ
For any x ∈ ℝ: x > 0 ⇒
{p ∈ ℚ: 0 < p)∩{q ∈ ℚ: q < x} is not empty.
{p ∈ ℚ: 0 < p)∩{q ∈ ℚ: q < x} ⊆ ℚ
there exists rational rₓ/sₓ:  0 < rₓ/sₓ < x
rₓ/sₓ  ∈  {p ∈ ℚ: 0 < p)∩{q ∈ ℚ: q < x}
for any x > 0
more.than.any.k<ℵ₀ unit.fractions
sit before x
among them are ⅟⌊(1+sₓ/rₓ)⌋ to ⅟⌊(k+1+sₓ/rₓ)⌋

Ax_def > 0: NUF(x_def) = ℵo is right.

∀x ∈ ℝ: x > 0 ⇒ NUF(x) = ℵ₀
ℝ is ℚ and points between non.∅ splits of ℚ

zero unit fractions sit before any x > 0

 So it is!
Therefore Ax > 0: NUF(x) = ℵo is wrong.

No. Your "therefore" is a quantifier shift.