Re: how

On 5/14/2024 8:47 AM, WM wrote:

Le 14/05/2024 à 13:58, Jim Burns a écrit :

On 5/13/2024 4:15 PM, WM wrote:

Then there would sit
ℵo unit fractions before any x > 0,
that is at x = 0.

>
No.

>
What is before any x > 0 and not 0?
Where do the first ℵo unit fractions sit?

| Tell her to find me an acre of land
| Parsley, sage, rosemary and thyme
| Between the salt water and the sea strands
| Then she'll be a true love of mine
|
-- "Scarborough Fair"

⎛ ¬∃x > 0: NUF(x) < ℵ₀
⎝ ∀x > 0: NUF(x) = ℵ₀

>
Don't repeat your claims.

It is a claim in a finite sequence of
only not.first.false claims.
Deleting the other claims from your post
doesn't change its being in sequence elsewhere.
Somewhere, a finite sequence of claims holds
only not.first.false claims.
Each of those claims is true, everywhere.

They are nonsensical unless
you can show the point x where
ℵ₀ unit fractions sit.
Where does NUF(x) change from 0 to ℵ₀?

In each open set O₀ holding 0,
NUF(x) changes from 0 to ℵ₀
However,
no point other than 0 (not.a.unit.fraction)
is in common with each open set O₀ holding 0
For example, consider x′ > 0
x′ is not.in open (-x′/2,x′/2)
but,
⎛ for each n ∈ ℕ
⎜ more.than.n unit fractions are in (-x′/2,x′/2)
⎝ 0 < ⅟⌊(2/x′+1+n)⌋ < ... < ⅟⌊(2/x′+1)⌋ < x′/2
ℵ₀.many unit fractions are in (-x′/2,x′/2)
x < -x′/2 ⇒ NUF(x) = 0
x′/2 < x  ⇒ NUF(x) = ℵ₀
NUF(x) changes from 0 to ℵ₀ in (-x′/2,x′/2)
but x′ ∉ (-x′/2,x′/2)
...and the same for each point > 0
You (WM) incorrectly think that that's incorrect,
I'd guess for the reason that
you incorrectly think that
a quantifier shift gives reliable results,
which it does not.