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

On 9/6/2024 4:52 PM, WM wrote:

On 06.09.2024 18:41, Jim Burns wrote:

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

>
Between 0 and your defined x or epsilon,
not between 0 and every possible x.

If x > 0  then there is ⅟⌊1+⅟x⌋  between 0 and x
⅟⌊1+⅟x⌋ is not between 0 and every possible x′ > 0
but ⅟⌊1+⅟x⌋ doesn't need to be.
The claim is "between 0 and x".
You (WM) think ⅟⌊1+⅟x⌋ needs every possible
because
you think a quantifier shift is reliable.
A quantifier shift is not reliable.

Between 0 and x
there are more.than.any.k<ℵ₀ unit fractions.

>
Betwee 0 and every epsilon you can define.

Between 0 and any epsilon satisfying my description.
I haven't made a claim for other epsilons.
⎛ n ∈ ℕ⁺ ⇔ n can be counted.to from 1
⎜⎛ ℕ⁺ is well.ordered
⎜⎜ min.ℕ⁺ = 1
⎜⎜∀n ∈ ℕ⁺: n-1 ∈ ℕ⁺ ⇔ n≠1
⎜⎝ ∀n ∈ ℕ⁺: n+1 ∈ ℕ⁺
⎜
⎜ ∀q: q ∈ ℚ⁺  ⇔
⎜ ∃j,k ∈ ℕ⁺: k⋅q = j
⎜
⎜ Splits.ℚ⁺ =
⎜ {S⊆ℚ⁺:∅≠Sᵉᵃᶜʰ<ₑₓᵢₛₜₛSᵉᵃᶜʰ<ᵉᵃᶜʰℚ⁺\S≠∅}
⎜
⎜ ∀r: r ∈ ℝ⁺  ⇔
⎝ ∃S' ∈ Splits.ℚ⁺: S' ᵉᵃᶜʰ< r ≤ᵉᵃᶜʰ ℚ⁺\S'
ε ∈ ℝ⁺

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 ?

>
ε = 1.

Not even close to first.
5⋅⌈½⌉ = 5
etc.

But that is not related to the topic.

The topic is: sets without minimums.
Such as {x ∈ ℝ: NUF(x) > 0}

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.

Can you answer
whether f(x) increases or decreases at 0?

>
It is constant 0.

f(x) = ⌊x⌋  or  f(x) = ⌈-x⌉
f(x) either increases or decreases at 0
f(0) = 0
f(x) is not constant.
Does f(x) increase at 0 or decrease at 0 ?
You can't answer.
You can only pretend I asked a different question.

No, you can't answer without information about
nearby points.

>
I have information about
nearby points on the negative axis.

Fascinating.
Information about f(x) = ⌊x⌋ or ⌈-x⌉ ?
Where did you get this information?

Only nearby points less than x are relevant
for judging about an increase in x.

Only points on the left.
And, if you step across the line,
only points on the before.right which are now.left.
Why?

Nearby points larger than x are irrelevant.

Why?
Another response like "Because you are stupid"
is you (WM) flailing.

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.

>
But there is an x immediately after 0.

There isn't in ℝ⁺
⎛ n ∈ ℕ⁺ ⇔ n can be counted.to from 1
⎜⎛ ℕ⁺ is well.ordered
⎜⎜ min.ℕ⁺ = 1
⎜⎜∀n ∈ ℕ⁺: n-1 ∈ ℕ⁺ ⇔ n≠1
⎜⎝ ∀n ∈ ℕ⁺: n+1 ∈ ℕ⁺
⎜
⎜ ∀q: q ∈ ℚ⁺  ⇔
⎜ ∃j,k ∈ ℕ⁺: k⋅q = j
⎜
⎜ Splits.ℚ⁺ =
⎜ {S⊆ℚ⁺:∅≠Sᵉᵃᶜʰ<ₑₓᵢₛₜₛSᵉᵃᶜʰ<ᵉᵃᶜʰℚ⁺\S≠∅}
⎜
⎜ ∀r: r ∈ ℝ⁺  ⇔
⎝ ∃S' ∈ Splits.ℚ⁺: S' ᵉᵃᶜʰ< r ≤ᵉᵃᶜʰ ℚ⁺\S'

Generalize.
What do we know about x ?
x > 0
What else?

>
We do not know anything.
But we know that
all unit fractions differ from each other.
That is sufficient to know that
NUF(x) increases by 1 only at any unit fraction.

NUF(x) never increases by 1.
If NUF(x) > 0 then NUF(x) > 1
If NUF(x) > 1 then NUF(x) > 2
If NUF(x) > 2 then NUF(x) > 3
  ...
If NUF(x) > 0 then,
for each k < ℵ₀  NUF(x) > k