Re: how

On 5/15/2024 9:30 AM, WM wrote:

Le 14/05/2024 à 22:07, Jim Burns a écrit :

NUF(x) changes from 0 to ℵ₀ in (-x′/2,x′/2)
but x′ ∉ (-x′/2,x′/2)

>
For every definable x'.

For each x' ∈ ℝ⁺ such that,
for each n ∈ ℕ
0 < ⅟⌊(2/x′+1+n)⌋ < ⅟⌊(2/x′+1)⌋ < x′/2
Thus,
for each x' ∈ ℝ⁺

...and the same for each point > 0

>
No.
∀n ∈ ℕ:
1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
NUF(x) cannot assume a value > 1
before it has assumed the value 1.

No.
No point x exists such that ⅟n < x
but not( ⅟(n+1) < ⅟n < x )
No point x exists such that NUF(x) = 1

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.

>
The function NUF(x) is defined at every x.
No further quantifier used.

You (WM) say
∀x > 0: NUF(x) = ℵ₀
is incorrect.
Which is to say, you (WM) say
∃x > 0: NUF(x) < ℵ₀
is correct.
| 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.
| Each non.∅ subset of ⅟ℕ∩(0,x′] is 2.ended,
| among them ⅟ℕ∩(0,x′] itself.
|
| Let ⅟G be the smallest in ⅟Ni(0,x']
| ⅟ℕ∩[⅟G,x′] = ⅟ℕ∩(0,x′]
| ⅟ℕ∩(0,⅟G) = ∅
|
| ⅟G ∈ ⅟ℕ
| G ∈ ℕ
| G can be counted.to from 0
| G+1 can be counted.to from G
| G+1 can be counted.to from 0 through G
| G+1 ∈ ℕ
| ⅟(G+1) ∈ ⅟ℕ
| ⅟(G+1) ∈ ⅟ℕ∩(0,⅟G)
| ⅟ℕ∩(0,⅟G) ≠ ∅
| Contradiction.
Therefore,
∃x > 0: NUF(x) < ℵ₀
is incorrect
∀x > 0: NUF(x) = ℵ₀
is correct.