Re: how

On 6/21/2024 2:59 AM, WM wrote:

Le 21/06/2024 à 02:18, Jim Burns a écrit :

On 6/19/2024 2:37 PM, WM wrote:

Le 18/06/2024 à 23:06, Jim Burns a écrit :

If the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
  does not hold a first element [in ⋂{Xᴬ⤾⁺¹₀}],
then the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
  is the empty set.

>
That is your big mistake!

>
For subsets of the union ⋃{⟨0…n⟩} of FISONs
only the empty set does not hold a first element.
[1]

>
That is true.
>
Since from every FISON F(n) = {1,2,3,...,n}
we know by definition that
it is neither sufficient nor necessary
to make the union of FISONs ℕ,
we can remove it from the union and
find
∪{F(1),F(2),F(3),...} = ℕ  ==>  ∪{ } = ℕ .

For each FISON, another FISON is proper superset.
∀ᶠⁱˢᵒⁿ⟨0…j⟩ ∃ᶠⁱˢᵒⁿ⟨0…k⟩≠⟨0…j⟩: ⟨0…j⟩ ⊂ ⟨0…k⟩
∀ᶠⁱˢᵒⁿ⟨0…j⟩ ∃ᶠⁱˢᵒⁿ⟨0…k⟩≠⟨0…j⟩: ⟨0…j⟩∪⟨0…k⟩ = ⟨0…k⟩
Thus, as you say:
∀ᶠⁱˢᵒⁿ⟨0…j⟩: ⋃{FISON} = ⋃({FISON}\{⟨0…j⟩})
However,
if all FISONs are removed,
no remaining FISON is a superset.
⋃({FISON}\{FISON}) = ⋃{}
⋃{} requires nuance.
Vacuously, everything is in each set in {}
Which makes ⋃{} the universal class.
Which is weird.
The convention is that ⋃{} = {}
But, either way, ⋃{} ≠ ℕ

Among the FISONs of ℕ there is not,
in any enumeration,
a first one that is required to yield
the union ℕ.

For each FISON, another FISON is proper superset.
∀ᶠⁱˢᵒⁿ⟨0…j⟩ ∃ᶠⁱˢᵒⁿ⟨0…k⟩≠⟨0…j⟩: ⟨0…j⟩ ⊂ ⟨0…k⟩
Therefore,
there is NO FISON superset all FISONs.
¬∃ᶠⁱˢᵒⁿ⟨0…k⟩ ∀ᶠⁱˢᵒⁿ⟨0…j⟩≠⟨0…k⟩: ⟨0…j⟩ ⊂ ⟨0…k⟩

Usually

Apparently, I am unusual.