Re: how

On 6/3/2024 3:50 PM, WM wrote:

Le 03/06/2024 à 20:34, Jim Burns a écrit :

On 6/3/2024 7:58 AM, WM wrote:

Le 03/06/2024 à 10:57, Jim Burns a écrit :

⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y   implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠xv : x<y

>
No this is not implied but
independently proven in Evidence for Dark Numbers,
prepublished chapter 4.2:
>
We assume that
all points on the [positive] real axis are fixed and
can be  subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.

2.
Or we can assume instead that
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
ℚ⁺ holds all.and.only ratios of numbers in ℕ⁺
ℝ⁺ holds all of ℚ⁺ and all.and.only
points x between open.foresplits Fₓ and ℚ⁺\Fₓ of ℚ⁺
with no points zero distance apart
and
⅟ℕ holds all.and.only reciprocals of numbers in ℕ⁺
ℝ⁺\⅟ℕ holds all.and only the others in ℝ⁺
and
ℝ⁺ is the positive real axis.

For visible numbers
we have two statements both of which are true:
[A]
There is no unit fraction smaller than
all positive non-unit fractions.
[B]
There is no positive non-unit fraction smaller than
all unit fractions

Under assumption (2.)
[A] and [B] are provable for all of ⅟ℕ and ℝ⁺\⅟ℕ

>
Hence assumption (2) contradicts logic.

Assumption (2.) describes
objects in our familiar arithmetic.
(2.)

ℕ⁺ holds all.and.only numbers countable.to
by.1 from.0

For each n in ℕ⁺
n countable.to from.0
n+1 is countable.to from.n
n+1 is countable.to from.0 through.n
n+1 is in ℕ⁺
n is not larger than all numbers in ℕ⁺
⎛ There is no number in ℕ⁺ larger than
⎝ all numbers in ℕ⁺
𝔼 is the subset of even numbers in ℕ⁺
𝕆 is the subset of odd numbers in ℕ⁺
⎛ There is no number in 𝔼 larger than
⎝ all numbers in 𝕆
⎛ There is no number in 𝕆 larger than
⎝ all numbers in 𝔼
(2.)

ℚ⁺ holds all.and.only ratios of numbers in ℕ⁺

For each q in ℚ⁺
exist j,k in ℕ⁺: j/k = q
k+1 in ℕ⁺
j/(k+1) in ℚ⁺: j/(k+1) < j/k = q
q is not smaller than all numbers in ℚ⁺
⎛ There is no number in ℚ⁺ smaller than
⎝ all numbers in ℚ⁺
For each u in ⅟ℕ ⊆ ℚ⁺
exist n in ℕ⁺: 1/n = u
n+1 in ℕ⁺
1/(n+1) in ⅟ℕ: 1/(n+1) < 1/n = u
u is not smaller than all numbers in ⅟ℕ
⎛ There is no number in ⅟ℕ smaller than
⎝ all numbers in ⅟ℕ
For each u in ⅟ℕ ⊆ ℚ⁺
exist n in ℕ⁺: 1/n = u
2/(2⋅n+1) in ℚ⁺\⅟ℕ: 2/(2⋅n+1) < 1/n = u
u is not smaller than all numbers in ℚ⁺\⅟ℕ
⎛ There is no number in ⅟ℕ smaller than
⎝ all numbers in ℚ⁺\⅟ℕ
For each q in ℚ⁺\⅟ℕ
exist j,k in ℕ⁺: j/k = q
exist n,r in ℕ⁺: n⋅j+r=k ∧ 0≤r<k
1/(n+1) in ⅟ℕ: 1/(n+1) < j/(n⋅j+r) = q
u is not smaller than all numbers in ⅟ℕ
⎛ There is no number in ℚ⁺\⅟ℕ smaller than
⎝ all numbers in ⅟ℕ
(2.)

ℝ⁺ holds all of ℚ⁺ and all.and.only
points x between open.foresplits Fₓ and ℚ⁺\Fₓ of ℚ⁺
with no points zero distance apart

For each x in ℝ⁺
exists open.foresplit F[x] = {q ∈ ℚ⁺: q<x}
exists open.foresplit F[x/2] = {q/2: q ∈ F[x]}
exists point x/2 in ℝ⁺ between F[x/2] and ℚ⁺\F[x/2]
ℝ⁺ ∋ x/2 < x
x is not smaller than all numbers in ℝ⁺
⎛ There is no number in ℝ⁺ smaller than
⎝ all numbers in ℝ⁺
For each u in ⅟ℕ
exists n in N+: 1/n = u
n+π ∉ ℕ⁺
⅟(n+π) ∉ ⅟ℕ
exists open.foresplit F[⅟(n+π)] = {q ∈ ℚ⁺: q<⅟(n+π)}
exists ⅟(n+π) in ℝ⁺ between F[⅟(n+π)] and ℚ⁺\F[⅟(n+π)]
ℝ⁺\⅟ℕ ∋ ⅟(n+π) < u
u is not smaller than all numbers in ℝ⁺\⅟ℕ
⎛ There is no number in ⅟ℕ smaller than
⎝ all numbers in ℝ⁺\⅟ℕ
For each x in ℝ⁺\⅟ℕ
exists open.foresplit F[x] = {q ∈ ℚ⁺: q<x}
F[x] ≠ ∅
exist j,k in ℕ⁺: j/k ∈ F[x]
exist n,r in ℕ⁺: n⋅j+r=k ∧ 0≤r<k
1/(n+1) in ⅟ℕ: 1/(n+1) < j/(n⋅j+r) < x
x is not smaller than all numbers in ⅟ℕ
⎛ There is no number in ℝ⁺\⅟ℕ smaller than
⎝ all numbers in ⅟ℕ

Only one of the two complementary sets
can and must contain the first point.

Why?
Responding "Logic" or "Mathematics" is dodging.

Not all subsets of unit fractions or
of non-unit fractions have two ends.

>
Pick a non.two.ended subset. 'Bye, Bob.
>

But this is dismissed by the fact that
the positive real axis and
all point sets in it
have an end at or before zero.

>
You're too late. Bob's gone.

>
Not in a mathematics based upon logic.

Why?
Responding "Logic" or "Mathematics" is dodging.