Re: There is a first/smallest integer (in Mückenland)

On 7/19/2024 9:53 AM, WM wrote:

Le 18/07/2024 à 23:35, Moebius a écrit :

Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

NUF(x) = aleph_0   for all x e IR, x > 0 ,

>
That means

>
that for each and every x e IR, x > 0
there are aleph_0 unit fractions which are <= x.

>
Then we can abbreviate each and every x > 0 by (0, oo).

Your "abbreviating" is a quantifier shift.
Making it less visible doesn't make it prove anything.
"Abbreviate" rock.paper.scissors and
you get wrong answers.
That method is unreliable.
----
For each A ⊆ ⅟ℕ∩(0,1]:  ∃u ∈ A ᴬ≤ u  ⇐  A ≠ {}
For each u ∈ ⅟ℕ∩(0,1]:  ⅟(1+⅟u) = max.⅟ℕ∩(0,u)
For each v ∈ ⅟ℕ∩(0,1]:  ⅟(-1+⅟v) = min.⅟ℕ∩(v,1]  ⇐  v < 1
The order of all unit fractions ⅟ℕ∩(0,1] is
well.ordered with step.down non.max.step.up
(max = 1)
There are ℵ₀.many unit.fractions in (0,1]
⎛ For each A ⊆ ⅟ℕ∩(0,x]:  ∃u ∈ A ᴬ≤ u  ⇐  A ≠ {}
⎜ because
⎜ for each A ⊆ ⅟ℕ∩(0,1]:  ∃u ∈ A ᴬ≤ u  ⇐  A ≠ {}
⎜ and
⎝ A ⊆ ⅟ℕ∩(0,x] ⊆ ⅟ℕ∩(0,1]
⎛ For each u ∈ ⅟ℕ∩(0,x]:  ⅟(1+⅟u) = max.⅟ℕ∩(0,u)
⎜ because
⎜ for each u ∈ ⅟ℕ∩(0,1]:  ⅟(1+⅟u) = max.⅟ℕ∩(0,u)
⎜ and
⎝ u ∈ ⅟ℕ∩(0,x] ⊆ ⅟ℕ∩(0,1]
⎛ For each v ∈ ⅟ℕ∩(0,x]:  ⅟(-1+⅟v) = min.⅟ℕ∩(v,x]  ⇐  v < ⅟⌈⅟x⌉
⎜ because
⎜ for each v ∈ ⅟ℕ∩(0,1]:  ⅟(-1+⅟v) = min.⅟ℕ∩(v,1]  ⇐  v < 1
⎜ and,
⎜ for v < ⅟⌈⅟x⌉:
⎜ ⅟ℕ∩(v,x] /= {}
⎝ min.⅟ℕ∩(v,x] = min.⅟ℕ∩(v,1]
For each A ⊆ ⅟ℕ∩(0,x]:  ∃u ∈ A ᴬ≤ u  ⇐  A ≠ {}
For each u ∈ ⅟ℕ∩(0,x]:  ⅟(1+⅟u) = max.⅟ℕ∩(0,u)
For each v ∈ ⅟ℕ∩(0,x]:  ⅟(-1+⅟v) = min.⅟ℕ∩(v,x]  ⇐  v < ⅟⌈⅟x⌉
The order of all unit fractions ⅟ℕ∩(0,x] is
well.ordered with step.down non.max.step.up
(max = ⅟⌈⅟x⌉)
There are ℵ₀.many unit.fractions in (0,x]
Infinite and humongous are different.

Ich glaube, das hat man Dir jetzt
schon so um die 500- bis 1000-mal erklärt, Mückenheim.
Hint: Ax > 0: E^aleph_0 u: ... <=/=> E^aleph_0 u: Ax > 0: ...

>
That is not claimed by quantifier shifting

Your "abbreviating" is a quantifier shift.
Making it less visible doesn't make it prove anything.
"Abbreviate" rock.paper.scissors and
you get wrong answers.
That method is unreliable.

but proven by the fact that
(0, oo) contains nothing but all x > 0.
Obviously it contains more than your x.
What could that be?

Proof by "obviously".
You depended upon the unreliable quantifier shift
and wishing upon a star ("obviously").
⅟ℕ∩(0,x] doesn't contain more than
max in each nonempty A
a step.down for each unit.fraction
a step.up for each non.max unit.fraction
That is sufficient for ℵ₀.many in ⅟ℕ∩(0,x]