Re: Replacement of Cardinality

On 8/19/2024 7:40 AM, WM wrote:

Le 17/08/2024 à 23:39, Jim Burns a écrit :

On 8/17/2024 9:28 AM, WM wrote:

Le 16/08/2024 à 19:39, Jim Burns a écrit :

no element of ℕᵈᵉᶠ is its upper.end,
because
for each diminishable k
diminishable k+1 disproves by counter.example
that k is the upper.end of ℕᵈᵉᶠ

>
SBZ(x) starts with 0 at 0 and increases,
but at no point x it increases by more than 1
because of
∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Therefore there is a smallest unit fractions
and vice versa a greatest natnumber.
What can't you understand?

>
How can ½⋅β
⎛ half the allegedly.positive greatest.lower.bound β of
⎝ visibleᵂᴹ.unit.fractions
be both lower.bound ( ½⋅β < β)
and not.lower.bound ( 2⋅β > ⅟k and ½⋅β > ¼⋅⅟k)
of the visibleᵂᴹ.unit.fractions?

>
Visible unit fractions have the lower bound 0.

Visibleᵂᴹ unit fractions have the greatest.lower.bound 0.
Otherwise, ½⋅β > 0 is contradictory.
⎛ not.bound 2⋅β > ⅟k
⎜ not.bound ½⋅β > ¼⋅⅟k
⎝ bound ½⋅β < β

Dark unit fractions have a smallest element.

Darkᵂᴹ unit.fractions are positive. Correct?
I don't want to alter your darkᵂᴹ

No smaller unit fractions is existing

Is that smallest darkᵂᴹ unit.fraction not a unit.fraction?
Is it not positive?
Each positive point,  thus,
  that smallest darkᵂᴹ unit.fraction
is not a lower.bound,  thus,
  has smaller unit.fractions, visibleᵂᴹ ones.
Are visibleᵂᴹ unit.fractions not existing?

No smaller unit fractions is existing
because no larger natnumber is existing.

For visibleᵂᴹ natural k
k∪{k} = k+1 disproves no larger.than.k existing.
Darkᵂᴹ natural 𝔊 has darkᵂᴹ unit.fraction ⅟𝔊
Darkᵂᴹ ⅟𝔊 is not.lower.bound of visiblesᵂᴹ ⅟k
Exists visibleᵂᴹ ⅟k < ⅟𝔊, thus k > 𝔊
For darkᵂᴹ natural 𝔊
k > 𝔊 disproves no larger.than.𝔊 existing.
Each natural, visibleᵂᴹ or darkᵂᴹ, is disproved
from no larger.than.it existing.
Each unit.fraction, visibleᵂᴹ or darkᵂᴹ, is disproved
from no smaller.than.it existing.