Re: Replacement of Cardinality

Le 04/08/2024 à 18:39, Jim Burns a écrit :

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

When all x are involved, the universal quantifier is usually not written.

 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

More of interest are these two claims which are not both true or both false:
For every x there is u < x.
There is u < x for every x.
The latter is close to my function:
There are NUF(x) u < x.

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

What claim you make is not of interest to me. I express that no u can be smaller than all x but that some u can be smaller than many x.
You express that for all x, there is a smaller u. Both are very different.

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

Not at all! Please spare these insults! Your claim concerns only definable x. For ℵo*2^ℵo undefinable points x it is wrong. My claim concerns all x. Regards, WM