Re: The set of necessary FISONs

On 2/5/2025 2:18 PM, WM wrote:

On 05.02.2025 19:43, Jim Burns wrote:

On 2/5/2025 12:32 PM, WM wrote:

On 05.02.2025 18:19, Jim Burns wrote:

Either way, the axioms do not create,
but only describe.

>
The axioms of
non-standard analysis or string theory
create.

 The axioms of
non-standard analysis or string theory
do not create, but only describe.

The axioms of Peano, Dedekind, Cantor,
Zermelo, Schmidt, v. Neumann, Lorenzen
concern all natural numbers
with no exception by induction.
By the same induction
I can remove all FISONs from U(F(n))
without changing the claimed union.

>
It might just matter what a natural number,
induction, a FISON, and a union are.

>
The axiom of induction:
∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
>
P(1): U(F(n) \ F(1)) = ℕ.
>
P(k): U(F(n) \ {F(1), F(2), ..., F(k)}) = ℕ
==>
P(k+1): U(F(n) \ {F(1), F(2), ..., F(k+1)}) = ℕ.

A description, not a magic spell.
...which could also be written...
∀P:ℕ₁→{⊤,⊥}:
P(1) ∧ ∀k∈ℕ₁:P(k)⇒P(k+1) ⇒ ∀n∈ℕ₁:P(n)
P(1) :⇔ ⋃({F(i):i∈ℕ₁}\{F(1)})=ℕ₁
⎛ P(k) :⇔ ⋃({F(i):i∈ℕ₁}\{F(i):i≤k∧i∈ℕ₁})=ℕ₁
⎜ ⇒
⎝ P(k+1) :⇔ ⋃({F(n):n∈ℕ₁}\{F(i):i≤k+1∧i∈ℕ₁})=ℕ₁
F(k) = {j∈ℕ₁:j≤k}

I can remove all FISONs from U(F(n))
without changing the claimed union.

What you (WM) mean by 'remove all...without changing..."
is that, by induction,
∀n∈ℕ₁:P(n)
which means
∀n∈ℕ₁:⋃({F(i):i∈ℕ₁}\{F(i):i≤n∧i∈ℕ₁})=ℕ₁
which, since
⋃({F(i):i∈ℕ₁}\{F(i):i≤n∧i∈ℕ₁}) = ∀n∈ℕ₁:⋃{F(i):n<i∧i∈ℕ₁}
means
∀n∈ℕ₁:⋃{F(i):n<i∧i∈ℕ₁}=ℕ₁
which means
∀n∈ℕ₁:⋃{{j∈ℕ₁:j≤i}:n<i∧i∈ℕ₁}=ℕ₁
which is true because                  [!]
∀ᴺ¹n: ∀j′∈ℕ₁: ∃ᴺ¹i′: n<i′ ∧ j′≤i′ ∧
  j′ ∈ ⋃{{j∈ℕ₁:j≤i}:n<i∧i∈ℕ₁}
  Proof: Let i′ = max{n+1,j′}. QED.
----
∀ᴺ¹n: ℕ₁ ⊆ ⋃{{j∈ℕ₁:j≤i}:n<i∧i∈ℕ₁}
And also,
∀ᴺ¹n: ℕ₁ ⊇ ⋃{{j∈ℕ₁:j≤i}:n<i∧i∈ℕ₁}
∀ᴺ¹n: ℕ₁ = ⋃{{j∈ℕ₁:j≤i}:n<i∧i∈ℕ₁}
∀ᴺ¹n: ⋃({F(i):i∈ℕ₁}\{F(i):i≤n∧i∈ℕ₁}) = ℕ₁
----
The essential part of that is
∀ᴺ¹n: ∃ᴺ¹i′: n<i′               [!]