Re: Replacement of Cardinality

On 7/29/2024 9:23 AM, WM wrote:

Le 28/07/2024 à 20:17, Jim Burns a écrit :

On 7/28/2024 8:17 AM, WM wrote:

Le 27/07/2024 à 19:34, Jim Burns a écrit :

On 7/26/2024 12:31 PM, WM wrote:

_The rule of subset_ proves that
every proper subset has less elements than its superset.

>

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

>
|ℕ| = ω-1 ∈ ℕ

>
No finite.ordinal is last.finite,
no visibleᵂᴹ finite.ordinal,
no darkᵂᴹ finite.ordinal.
In particular, no finite.ordinal is ω-1

>
NUF(x) cannot grow by more than 1 at any x.

NUF(x) = |⅟ℕ∩(0,x]|
NUF(x) cannot grow by more than 1 at any x > 0
¬(0 > 0)
⅟ℕ∩(0,x] has
each non.{}.subset maximummed
each unit.fraction down.stepped
each non.maximum up.stepped
and therefore
ℵ₀.many unit.fractions
x > 0  ⇒  NUF(x) = ℵ₀

Therefore there is a first step.

0 < ⅟⌈1+⅟x⌉ < ⅟⌈⅟x⌉ ≤ x
⅟⌈⅟x⌉ ∈ ⅟ℕ∩(0,x]
⅟⌈1+⅟x⌉ ∈ ⅟ℕ∩(0,x]
NUF(x) = |⅟ℕ∩(0,x]| ≠ 1

First unit fraction implies last natural number.

Ordinal λ = the set of ordinals < λ
λ = [0,λ)ᴼʳᵈ
( k is a finite.ordinal
⇔
( each non.{}.subset of [0,k)ᴼʳᵈ is 2.ended
_Finite doesn't need to be small_
( k is a finite.ordinal
⇔
⎛ [0,k)ᴼʳᵈ is 2.ended  and
⎝ ∀j ∈ (0,k)ᴼʳᵈ:  [0,j)ᴼʳᵈ is 2.ended
if
( k is a finite.ordinal
then
( k+1 is a finite ordinal
...because
if
⎛ [0,k)ᴼʳᵈ is 2.ended  and
⎝ ∀j ∈ (0,k)ᴼʳᵈ:  [0,j)ᴼʳᵈ is 2.ended
then
⎛ [0,k+1)ᴼʳᵈ is 2.ended  and
⎝ ∀j ∈ (0,k+1)ᴼʳᵈ:  [0,j)ᴼʳᵈ is 2.ended
ω is defined to be the first transfinite.ordinal.
ω is not a reallyreallyreally big finite.ordinal.
ω = [0,ω)ᴼʳᵈ
if ω-1 exists
then
⎛ ω-1 = max.[0,ω)ᴼʳᵈ  and
⎜⎛ [0,ω)ᴼʳᵈ is 2.ended  and
⎜⎝ ∀j ∈ (0,ω)ᴼʳᵈ:  [0,j)ᴼʳᵈ is 2.ended
⎜ and
⎝ ω is finite.
However,
ω is not finite, not even a reallyreallyreally big one.
ω is transfinite (the first).
[0,ω)ᴼʳᵈ isn't 2.ended.
ω-1 doesn't exist.

Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1

>
Don't claim.

Before the first star, no stars existed.
Before the first mammal, no mammal existed.
Before the first infinite.ordinal, no infinite ordinal exists.
What ω means is the first infinite ordinal.

Explain how you imagine
the positions of the unit fractions such that
NUF does not have a first step.

ℚ the rationals are fractal.
For each rational p/q in [0,⅟n]
there is rational n⋅p/q in [0,1]
Zoom from [0,1] in to [0,⅟n]
[0,⅟n] looks just like [0,1]
Zoom from [0,⅟n] in to [0,⅟n²]
[0,⅟n²] looks just like [0,⅟n] and
[0,⅟n] looks just like [0,1]
No zoom.level ⅟nₓ exists such that
[0,⅟nₓ] does NOT look like [0,1]
No zoom.level ⅟nₓ exists such that
some unit.fraction.free gap opens.
[0,⅟nₓ] looks like [0,1] and
[0,1] isn't unit.fraction.free.
No zoom.level ⅟nₓ exists such that
[0,⅟nₓ] holds fewer unit.fractions than [0,1]
No x > 0 exists such that
NUF(x) ≠ NUF(1) = ℵ₀
----
A lagniappe.
ℚ the rationals are fractal.
ℝ the reals are fractal AND each split is situated in ℝ:
For each ℝ.split F′ ᵃˡˡ<ᵃˡˡ H′
some x′ ∈ ℝ is last.in.F′ or first.in.H′
With each zoom.level ⅟n
x′ is at the split F′ ᵃˡˡ<ᵃˡˡ H′