Re: How many different unit fractions are lessorequal than all unit fractions?

On 9/11/2024 10:48 AM, WM wrote:

On 10.09.2024 23:12, Jim Burns wrote:

On 9/10/2024 12:16 PM, WM wrote:

On 09.09.2024 21:43, Jim Burns wrote:

The unit fractions
  before x > 0 and before x′ > 0
vary.
For both x and x′, they are ℵ₀.many,
but not the same ℵ₀.many.

>
The same ℵ₀ many are existing in both cases,

>
No.
They are not the same for ⌊⅟x⌋ < ⌊⅟x′⌋

>
The unit fractions are fixed at the real line.
If you cut some more or some less by x than by x'
is irrelevant for the fact that
the same ℵo unit fractions remain in both cases.

⅟⌊k+⅟x⌋ ⟷ ⅟⌊k+⅟x′⌋
is a bijection between different visibleᵂᴹ unit.fractions
⅟⌊k+⅟x⌋ ≠ ⅟⌊k+⅟x′⌋
⅟ℕ⁺∩(0,x) the visibleᵂᴹ unit.fractions < x
is bijectible with
⅟ℕ⁺∩(0,x′) the visibleᵂᴹ unit.fractions < x′
but they hold different unit.fractions.
Also, ⅟ℕ⁺∩(0,x) is a proper superset of ⅟ℕ⁺∩(0,x′)
which can't be true for bijectible finite sets.
Conclusion: ⅟ℕ⁺∩(0,x) and ⅟ℕ⁺∩(0,x′) aren't finite sets.
Whether darkᵂᴹ exist or not.exist,
those facts about sets of visibleᵂᴹ unit fractions
do not change.
⎛ There is a separate argument that
⎜ a positive greatest.lower.bound β of ⅟ℕᵈᵉᶠ
⎜ requires that
⎜ ½.β is undercut by a visible unit.fraction and
⎜ ½.β is NOT undercut by a visible unit.fraction
⎜ It's impossible for both to be true.
⎜
⎜ Therefore,
⎜ it is impossible for
⎝ darkᵂᴹ points to be between 0 and ⅟ℕᵈᵉᶠ
With or without darkᵂᴹ points,
∀ᴿx > 0: NUF(x) ≥ NUFᵈᵉᶠ(x) = ℵ₀

0, ..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.

0 is not a unit.fraction.
..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.
holds only visibleᵂᴹ unit fractions
(darkᵂᴹ not.exist)
For each visible unit fraction,
there is a counter.example to
the claim that it is first.
Anything not.in
..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.
is not.first.in
..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.
Therefore,
no first exists of
..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.
..., 1/100, x', 1/99, ..., 1/75, x, 1/74, ..., 1/1.
is infinite.

_Minimum distances of countably.many.points does not exist_

>
Every uncountable set has a countable subset,
in fact uncoubtably many.
Therefore (0, d) c (0, x).

(0,d) is an interval as well as a set.
(0,d) := {y ∈ ℝ: 0 < y < d}
y ⟷ d⋅y
is a bijection between (0,1) and (0,d)
|(0,1)| = |(0,d)|
Also, (0,1) is a proper superset of (0,d)
which can't be true for bijectible finite sets.
Conclusion: (0,1) and (0,d) aren't finite sets.

(0, d) is coubtable and contains at most one unit fraction.

No.
0 < ... < ⅟⌊2+⅟d⌋ < ⅟⌊1+⅟d⌋ < d