Re: how

On 5/16/2024 2:21 PM, WM wrote:

Le 15/05/2024 à 19:30, Jim Burns a écrit :

| Assume
| ∃x > 0: NUF(x) < ℵ₀
| is correct.
|
| Assume
| x′ exists such that |⅟ℕ∩(0,x′]| < ℵ₀
|
| ⅟ℕ∩(0,x′] is totally.ordered and finite.

>
Not in the dark numbers.

Not in the visibleᵂᴹ or darkᵂᴹ numbers.
Thus
∃x > 0: NUF(x) < ℵ₀
is NOT correct.
∀x > 0: NUF(x) = ℵ₀
IS correct.

∀x > 0: NUF(x) = ℵ₀
is correct.

>
Tell me where the first point
loaded with ℵ₀ unit fractions is.

No x′ > 0 exists such that
⅟ℕ∩(0,x′] is totally.ordered and finite.
¬∃x > 0: NUF(x) < ℵ₀
∀x > 0: NUF(x) = ℵ₀
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| Between the salt water and the sea strands
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|
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