Re: how

On 6/23/2024 4:04 PM, WM wrote:

Le 23/06/2024 à 19:58, Jim Burns a écrit :

U{FISON}  is the union of the set of
both necessary and unnecessary FISONs

>
The unnecessary FISONs can be removed.

Your definition of 'unnecessary' is that
an unnecessaryᵂᴹ FISON when removed
leaves the union of remaining FISONs unchanged.
One unnecessaryᵂᴹ FISON can be removed.
One.
{FISON} is inductive and well.ordered.
Because well.ordered,
Any necessaryᵂᴹ FISON in {FISON}
is last in {FISON}
Because inductive,
No FISON is last in {FISON}
Any _one_ FISON can be removed and
leave the union of remaining FISONs unchanged.
Any _one_ FISON is unnecessaryᵂᴹ
Removing all of those which
singly don't change the union
changes the union, ⋃{}≠ℕ
It is not a contradiction.
Compare to:
Suppose I have 5 elephants, and
an unnecessaryᴶᴮ elephant is one when removed
leaves 4 elephants.
Any _one_ elephant can be removed and
leave 4 elephants.
Any _one_ elephant is unnecessaryᴶᴮ
Removing all which
singly leave 4
doesn't leave 4
It is not a contradiction.
It is arithmetic.

The union cannot be larger than all its FISONs.

>
The union cannot be smaller than any FISON.

>
Agreed.

Each FISON is smaller than another FISON,

Therefore the union is a FISON.

Therefore the union is NOT a FISON.

The sequence of FISONs is potentially infinite.

{FISON} is inductive and well.ordered.

There is no last one. The end is evolving.

There is no last one.
The end is not.existing.

It runs through ℕ
but never covers a substantial part.

Related:
∀ᴿx>0: NUF(x) = ℵ₀

Yes,
the set of FISONs necessary to yield ℕ is empty.

>
That proves that
ℕ is more than all FISONs.

It equally.well proves that
5 elephants are more than 5 elephants.

According to Cantor's theorem B
the set of necesary FISONs must have
a smallest element.

The empty set doesn't have a smallest element.
FISON.union ⋃{FISON} and minimal.inductive ⋂{inductive}
are both inductive and both well.ordered
⋃{FISON} holds NO FIRST not.in ⋂{inductive}
⋃{FISON} holds NONE not.in ⋂{inductive}
⋂{inductive} holds NO FIRST not.in ⋃{FISON}
⋂{inductive} holds NONE not.in ⋃{FISON}
⋃{FISON} = ⋂{inductive}

This proves that ℕ is not the union of FISONs.

Is ℕᵂᴹ inductive and well.ordered?
If yes,
ℕᵂᴹ = ⋃{FISON} = ⋂{inductive}