Re: Replacement of Cardinality

On 8/4/2024 11:29 AM, WM wrote:

Le 03/08/2024 à 21:54, Jim Burns a écrit :

On 8/3/2024 10:23 AM, WM wrote:

I recognized lately that you use
the wrong definition of NUF.
>
Here is the correct definition:
There exist NUF(x) unit fractions u, such that
for all y >= x: u < y.

>

Here is an equivalent definition:
There exist NUF(x) unit fractions u, such that
u < x
>

Note that the order is ∃ u ∀ y.

>
The order is ∀x ∃u ∀y

The order of the claim which you (WM) address
in an attempt to "prove" dark numbers is
∀ᴿx > 0:
∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
∀ᴿy ≥ x:
y >ᵉᵃᶜʰ U
That claim and the following claim are
either both true or both false.
∀ᴿx > 0:
∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
x >ᵉᵃᶜʰ U

∃u ∀x ∀y is an unreliable quantifier shift.

>
It is not a shift but it is the definition of NUF.

Your recently corrected definition of NUF is
NUF(x) =
|{u ∈ ⅟ℕ: ∀ᴿy ≥ x: y > u}|
That definition is equivalent to
NUF(x) =
|{u ∈ ⅟ℕ: x > u}|
Note that,
for x > 0, {u ∈ ⅟ℕ: x > u}
is maximummed and down.stepped and non.max.up.stepped.
For x > 0:  |{u ∈ ⅟ℕ: x > u}| = ℵ₀
The claim you (WM) use
∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
∀ᴿx > 0:
∀ᴿy ≥ x:
y >ᵉᵃᶜʰ U
is an unreliable quantifier shift from
the claim we make
∀ᴿx > 0:
∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
∀ᴿy ≥ x:
y >ᵉᵃᶜʰ U

It excludes that ∃u ∀x>0: u < x,

Only you (WM) think that ∃u ∀x>0: u < x
follows from ∀x>0 ∃u: u < x,