Re: The set of necessary FISONs

On 2/2/2025 6:25 AM, WM wrote:

On 01.02.2025 20:21, Jim Burns wrote:

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

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).

>
But all F(n) can be discarded
without changing the union.

Yes, because
none in linearly.ordered {FISON} is last.
F(0)∪F(1)∪...∪F(n-1)∪F(n)∪F(n+1)∪...  =
F(0)∪F(1)∪...∪F(n-1)∪F(n+1)∪...

F(1) can be discarded.
If F(n) can be discarded, then F(n+1) can be discarded.
>
Note:
Mathematical induction is a method for proving that
a statement P(n) is true for every natural number n
that is, that
the infinitely many cases P(0),P(1),P(2),P(3),...
all hold. [Wikipedia]

For each k ∈ ⋃{FISON}:
⋃{FISON} = ⋃({FISON}\{F(k)})
P(k)  :⇔  ⋃{FISON}=⋃({FISON}\{F(k)})
For each k ∈ ⋃{FISON}:  P(k)
Because
none in linearly.ordered {FISON} is last.
{i∈⋃{FISON}: P(i)}  =  ⋃{FISON}

Therefore if U(F(n)) = ℕ, then  { } = ℕ

If we define  ℕ := ⋃{FISON}
then  ⋃{FISON} = ℕ
If {0,1} ∈ {FISON}
then  {} ≠ ⋃{FISON} = ℕ

Therefore if U(F(n)) = ℕ, then  { } = ℕ

You (WM) don't say why that should follow.
Most likely, it's because you deny that
none in linearly.ordered {FISON} is last.
ᵂᴹ⎛ ∀k ∈ ⋃{FISON}:
ᵂᴹ⎜ ⋃( {FISON}\{F(k)} ) =
ᵂᴹ⎜ ⋃( {FISON}\⋃{F(j):j≤k} )
ᵂᴹ⎜
ᵂᴹ⎜⎛ Assume
ᵂᴹ⎜⎜ ω-1 is last.
ᵂᴹ⎜⎜ ∀j ∈ ⋃{FISON}:  j ≤ ω-1 ∈ ⋃{FISON}
ᵂᴹ⎜⎜
ᵂᴹ⎜⎜ {F(j):j≤ω-1} = {FISON}
ᵂᴹ⎜⎜
ᵂᴹ⎜⎜ U( {FISON}\{F(ω-1)} ) =
ᵂᴹ⎜⎜ U( {FISON}\U{F(j):j≤ω-1} ) =
ᵂᴹ⎜⎜ U( {FISON}\U{FISON} ) =
ᵂᴹ⎜⎜ U{FISON}\U{FISON} =
ᵂᴹ⎜⎝ {}
ᵂᴹ⎜
ᵂᴹ⎜ From above,
ᵂᴹ⎜ U( {FISON}\{F(ω-1)} ) = ℕ
ᵂᴹ⎜
ᵂᴹ⎝ {} = ℕ
The problem with that reasoning,
which is better than any reasoning you (WM) offer,
(AKA nothing)
is that
each finiteᵒᵘʳ initial segment F(k) of ⋃{F(n)}
can grow¹ to another initial segment F(k+1)
-- because it's finiteᵒᵘʳ --
which is also finiteᵒᵘʳ, and is larger than F(k)
-- because they're finiteᵒᵘʳ --
and is not larger than ⋃{F(n)}
-- because F(k+1) ⊆ ⋃{F(n)} --
No FISON is last in {FISON}
-- because FISONs are finiteᵒᵘʳ --
----
|A| < |B|  ⇔  |Aᵃ| < |Bᵇ|
Aᵃ = A∪{a} ≠ A
Bᵇ = B∪{b} ≠ B
¬(|A| > |Aᵃ|)
|A| < |Aᵃ|  ∨  |A| = |Aᵃ|
Let B = Aᵃ
|A| < |Aᵃ|  ⇔  |Aᵃ| < |(Aᵃ)ᵇ|
|A| < |Aᵃ| < |(Aᵃ)ᵇ|  ∨  |A| = |Aᵃ| = |(Aᵃ)ᵇ|
Either
Aᵃ is both.growable¹.and.shrinkable¹
or
Aᵃ is both.ungrowable¹.and.unshrinkable¹.
Therefore,
there can't be both
both.growable¹.and.shrinkable¹ ω-1  and
both.ungrowable¹.and.unshrinkable¹ ω
finiteᵒᵘʳ == both.growable¹.and.shrinkable¹
infiniteᵒᵘʳ == both.ungrowable¹.and.unshrinkable¹
there can't be both
finiteᵒᵘʳ ω-1  and infiniteᵒᵘʳ ω