Re: The set of necessary FISONs

On 2/1/2025 7:56 AM, WM wrote:

On 31.01.2025 19:34, Jim Burns wrote:

On 1/30/2025 5:32 PM, WM wrote:

ℕ holds only completely.uselessᵂᴹ numbers.
ℕ isn't what you (WM) think it is.
That's not a logic.problem.
That's a you.problem.

>
There is the assumption that
a set with U(F(n)) = ℕ exists.

The you.problem is that
there's what.you.think.there.is
and then there's what.there.is
finiteᵂᴹ     infiniteᵂᴹ   wrongᵂᴹ.and.matheologicalᵂᴹ
   ↕            ↕                ↕
finiteᵒᵘʳ   more.finiteᵒᵘʳ   infiniteᵒᵘʳ
What.you.think.there.is is
an upper.edgeᵂᴹ to finiteᵂᴹ
where one can imagine stepping to infiniteᵂᴹ
although one can't in practice, because of the darkᵂᴹ
What.there.is and what there.only.is  is
sets  both.growable¹.and.shrinkable¹  and
sets  both.ungrowable¹.and.unshrinkable¹.
Finiteᵒᵘʳ is both.growable¹.and.shrinkable¹.
Infiniteᵒᵘʳ is both.ungrowable¹.and.unshrinkable¹.
No upper.edgeᵒᵘʳ exists to finiteᵒᵘʳ.
Your darkᵂᴹ upper.edgeᵂᴹ is between
finiteᵒᵘʳ and more.finiteᵒᵘʳ.
Everywhere in finiteᵒᵘʳ,
the only steps are from growable¹ to shrinkable¹
or from shrinkable¹ to growable¹.
There can't be a set which was grown¹ to
which can't be shrunk¹ from.
But growable¹ and shrinkable¹ march in lockstep.
Everywhere in finiteᵒᵘʳ,
the only steps are
from both.growable¹.and.shrinkable¹
to both.growable¹.and.shrinkable¹,
the only steps are
from finiteᵒᵘʳ to finiteᵒᵘʳ.
A finiteᵒᵘʳ set has fuller¹ sets,
but no fuller¹ infiniteᵒᵘʳ sets.
No upper.edgeᵒᵘʳ exists to finiteᵒᵘʳ.
Darkᵂᴹ or visibleᵂᴹ,
no upper.edgeᵒᵘʳ exists to finiteᵒᵘʳ.
That is because
finiteᵒᵘʳ is both.growable¹.and.shrinkable¹,
and growable¹ and shrinkable¹ march in lockstep.
It's also true that
infiniteᵒᵘʳ is both.ungrowable¹.and.unshrinkable¹,
and ungrowable¹ and unshrinkable¹ march in lockstep.
By similar reasoning,
an infiniteᵒᵘʳ set has emptier¹ sets,
but no emptier¹ finiteᵒᵘʳ sets.
No lower.edgeᵒᵘʳ exists to infiniteᵒᵘʳ.
I imagine hearing
( Ah HAH! What about first.infiniteᵒᵘʳ ω ?
Last.finiteᵒᵘʳ ω-1 a step from ω
can't exist.
There is no 'there' there.
That's in the ordinals.
Sets of ordinals are first.holders or empty.
As first among the infiniteᵒᵘʳ, ω must exist.
But then ω-1 can't exist.
Consider a model of Robinson arithmetic,
with non.standard natural numbers: ℕ∪ᑉℤꜝ
Each n ∈ ℕ is finiteᵒᵘʳ.
Each z ∈ ℤꜝ is infiniteᵒᵘʳ.
ℕ  is well.ordered, but ℤꜝ isn't.
For each infiniteᵒᵘʳ z ∈ ℤꜝ, there is a "there" there,
immediately before that z.
Even here, though,
there is no step from infiniteᵒᵘʳ to finiteᵒᵘʳ,
only steps from ℤꜝ to ℤꜝ
There can't be a step from ℤꜝ to ℕ.
This is more permanent than
"We have decided these are ordinals", or
"We have decided this is Robinson arithmetic".
It is because
growable¹ and shrinkable¹ march in lockstep,
something which can't be decided to be otherwise.

There is the assumption that
a set with U(F(n)) = ℕ exists.
Without changing the union
we can remove every element by induction.
No element remains.
The set does not exist.

Each finiteᵒᵘʳ initial segment F(k) of ⋃{F(n)}
can grow¹ to another initial segment F(k+1)
which is also finiteᵒᵘʳ, and is larger than F(k),
and is not larger than ⋃{F(n)}
{F(n}} holds each finiteᵒᵘʳ initial segment F(k)
⋃{F(n)} is larger than each F(k).
The sum of any two finiteᵒᵘʳ initial segments F(j),F(k)
is also a finiteᵒᵘʳ initial segment F(j+k)
For any finiteᵒᵘʳ initial segment F(j)
the end.segment ⋃{F(n)}\F(j)
is larger than each F(j+k)\F(j)
is larger than each F(k)
is not larger than minimal ⋃{F(n)}
is not smaller than superset ⋃{F(n)} ⊇ ⋃{F(n)}\F(j)
For any finiteᵒᵘʳ initial segment F(j)
the end.segment ⋃{F(n)}\F(j)
is the size of ⋃{F(n)}