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.

Thank you.
Also,
⋃{⟨0…n⟩} is inductive.
( ⋃{⟨0…n⟩} )ᴬ⤾⁺¹₀
As it is with all inductive sets,
inductive FISON.union ⋃{⟨0…n⟩} is a superset of
the minimal inductive ⋂{Xᴬ⤾⁺¹₀}
⋃{⟨0…n⟩}  ⊇  ⋂{Xᴬ⤾⁺¹₀}
Each nonempty ⋂{Xᴬ⤾⁺¹₀}.subset is
a nonempty ⋃{⟨0…n⟩}.subset
Being a nonempty ⋃{⟨0…n⟩}.subset, it holds a first element.
Thus, _also_
for subsets of the minimal inductive ⋂{Xᴬ⤾⁺¹₀}
only the empty set does not hold a first element.
And also the minimal inductive is inductive.
( ⋂{Xᴬ⤾⁺¹₀} )ᴬ⤾⁺¹₀
tl;dr
⋃{⟨0…n⟩} and ⋂{Xᴬ⤾⁺¹₀} are
both inductive and both well.ordered.
----
 From that, it follows that
⋃{⟨0…n⟩}  =  ⋂{Xᴬ⤾⁺¹₀}
Consider:
There is NO FIRST in ⋃{⟨0…n⟩} not.in ⋂{Xᴬ⤾⁺¹₀}
because ⋂{Xᴬ⤾⁺¹₀} is inductive, so
there is NO j ∉ ⋂{Xᴬ⤾⁺¹₀}  ∧  j⁻¹ ∈ ⋂{Xᴬ⤾⁺¹₀}
There is NO FIRST in ⋃{⟨0…n⟩} not.in ⋂{Xᴬ⤾⁺¹₀}
There is NONE in ⋃{⟨0…n⟩} not.in ⋂{Xᴬ⤾⁺¹₀}
By a similar argument,
there is NO FIRST in ⋂{Xᴬ⤾⁺¹₀} not.in ⋃{⟨0…n⟩}
There is NONE in ⋂{Xᴬ⤾⁺¹₀} not.in ⋃{⟨0…n⟩}
∀j: j ∈ ⋃{⟨0…n⟩}  ⇔  j ∈ ⋂{Xᴬ⤾⁺¹₀}
⋃{⟨0…n⟩}  =  ⋂{Xᴬ⤾⁺¹₀}
----
{0<} the 0.followers in ⋂{Xᴬ⤾⁺¹₀} are ℵ₀.many.
There is NO FIRST jₓ in ⋂{Xᴬ⤾⁺¹₀} which has
fewer.than.ℵ₀.many jₓ.followers {jₓ<} in ⋂{Xᴬ⤾⁺¹₀}
because
sⱼₓ(i) = i⁺¹
Sⱼₓ:{jₓ⁻¹<}⇉{jₓ<}: 1.to.1
|jₓ⁻¹<| ≤ |jₓ<|
idⱼₓ(i) = i
idⱼₓ:{jₓ<}⇉{jₓ⁻¹<}: 1.to.1
|jₓ<| ≤ |jₓ⁻¹<|
|jₓ⁻¹<| = |jₓ<|
But, for FIRST jₓ in ⋂{Xᴬ⤾⁺¹₀}
|jₓ<| < ℵ₀
|jₓ⁻¹<| = ℵ₀
|jₓ⁻¹<| ≠ |jₓ<|
Contradiction.
There is NO FIRST jₓ in ⋂{Xᴬ⤾⁺¹₀}  and
there is NO jₓ in ⋂{Xᴬ⤾⁺¹₀} which has
fewer.than.ℵ₀.many jₓ.followers {jₓ<} in ⋂{Xᴬ⤾⁺¹₀}

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 .
>
Among the FISONs of ℕ there is not,
in any enumeration,
a first one that is required to yield
the union ℕ.
>
Usually this is apologized by the fact,
that even in
{1, 2} ∪ {2, 3} ∪ {3, 1} = {1, 2, 3}
 (*)
>
it is impossible to find a first set which
cannot be omitted from the union
to yield {1, 2, 3}.
But this argument fails.
It is not a set of sets which is subject to
Cantor's theorem B
(every embodiment of different numbers of
the first and the second number class
has a smallest number)
but only every set of ordinal numbers.
Therefore
we always have to enumerate the sets.
In case of FISONs this is simple.
We apply the natural order:
{1,2,3,...,n} --> n.
Of course
every other enumeration would also do.
In case of the sets (*)
we can use the written order from left to right.
Then the first set not to be omitted is {2, 3}
because after having omitted {1, 2} already,
2 would then be missing in the union.
Every other order is possible and
has a first set which cannot be omitted.