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

On 9/6/2024 8:12 AM, WM wrote:

On 05.09.2024 21:57, Jim Burns wrote:

On 9/5/2024 9:59 AM, WM wrote:

NUF(x) increases from 0 to more.

>
...at 0.

>
No.
NUF(x) counts the unit fractions between 0 and x.
Between 0 and 0 there is no unit fraction.
NUF(0) = 0.

Between 0 and x
there are more than 0 unit fractions
there is more than 1 unit fraction
there are more than 2 unit fractions
there are more than 3 unit fractions
...
0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x
Between 0 and x
there are more.than.any.k<ℵ₀ unit fractions.
0 < ... < ⅟⌊k+1+⅟x⌋ < ... < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x

It cannot increase to 2 or more
before having accepted 1.

>
NUFᵈᵉᶠ(x) cannot increase to 2
without having already been ≥ 2

>
Impossible.
NUF(0) = 0.
There is a first increase in linear order.

∀ᴿx ∈ (-1,0]:  ⌈x⌉ = 0
∀ᴿx ∈ (0,1]:  ⌈x⌉ = 1
Is there a first ε at which 5⋅⌈ε⌉ = 5 ?
Is there a first ε at which 5⋅⌈ε⌉ = 1 ?
Why is there?

It cannot increase to ℵo without
having accepted 1, 2, 3, ...

>
x > 0  ⇒
0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x

>
0 is smaller than all that.
Therefore there is no increase at 0.

Increases and decreases involve nearby points
from which increases and decreases increase and decrease.
∀ᴿx ∈ [-1,0):  ⌊x⌋ = -1
∀ᴿx ∈ [0,1):  ⌊x⌋ = 0
∀ᴿx ∈ [-1,0):  ⌈-x⌉ = 1
∀ᴿx ∈ [0,1):  ⌈-x⌉ = 0
f(x) = ⌊x⌋  or  f(x) = ⌈-x⌉
f(x) either increases or decreases at 0
f(0) = 0
Can you answer
whether f(x) increases or decreases at 0?
No, you can't answer without information about
nearby points.

x is larger than all that.

"All that" are more.than.any.k<ℵ₀ unit.fractions.
NUF(x) = ℵ₀

Therefore your x is not the least one posiible.

Yes.
x is NOT the first point > 0 after
more.than.any.k<ℵ₀ unit.fractions.
Generalize.
What do we know about x ?
x > 0
What else?