Re: The set of necessary FISONs

On 2/12/2025 4:19 AM, WM wrote:

On 12.02.2025 01:38, Jim Burns wrote:

On 2/11/2025 2:23 PM, WM wrote:

On 11.02.2025 18:42, Jim Burns wrote:

On 2/11/2025 4:31 AM, WM wrote:

The set F of FISONs which can be removed
without changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
This is proven by just the same induction
as Zermelo proves his infinite set Z.
Either you accept both proofs or none.

>
What is this "both proofs"?

>
One is Zermelo's proof by induction
that there is an infiite set Z.

https://en.wikipedia.org/wiki/Zermelo_set_theory
The proof of Z you refer to is the bare assertion
that inductiveᶻ Z exists, AXIOM VII "Infinity".
  Z exists such that Z∋{} ∧ ∀a∈Z∋{a}
That bare assertion of an axiom is a proof,
a very short proof,
but not a proof by induction.
In Zermelo's domain:
⎛
⎜ By VII
⎜ Inductiveᶻ Z exists:
⎜ Z∋{} ∧ ∀a∈Z∋{a}
⎜
⎜⎛ Define
⎜⎜ 0 := {}
⎜⎜ a+1 := {a}
⎜⎜
⎜⎝ Z∋0 ∧ ∀a∈Z∋a+1
⎜
⎜ By VI
⎜ Powerset 𝒫(Z) = {S ⊆ Z} of Z exists
⎜
⎜ By III
⎜ Set 𝒫ⁱⁿᵈ(Z) of inductiveᶻ subsets of Z exists
⎜ 𝒫ⁱⁿᵈ(Z) = {S ∈ 𝒫ⁱⁿᵈ(Z): inductiveᶻ S }
⎜
⎜ By III
⎜ Intersection ⋂𝒫ⁱⁿᵈ(Z) of inductiveᶻ subsets of Z exists
⎝ ⋂𝒫ⁱⁿᵈ(Z) = {a ∈ Z: AS ∈ 𝒫ⁱⁿᵈ(Z): S ∋ a}
That is a proof that ⋂𝒫ⁱⁿᵈ(Z) exists,
but not a proof by induction.
⋂𝒫ⁱⁿᵈ(Z) exists,
is the same set for different inductive sets Z,
is infinite,
is inductive, and
is minimal.inductive.
⋂𝒫ⁱⁿᵈ(Z) is one example of what we mean by ℕ
There are other examples.
Each example (each model) satisfies defining claims,
primarily the claim that it is minimal.inductive.
Whatever we prove from the assumption of
the defining claims is true of
anything which satisfies those defining claims,
and that anything can be any model of ℕ
Because ⋂𝒫ⁱⁿᵈ(Z) is minimal.inductive,
P(0) ∧ ∀k∈⋂𝒫ⁱⁿᵈ(Z): P(k)⇒P(k+1)
⇒ ∀n∈⋂𝒫ⁱⁿᵈ(Z): P(n)
Also, with ℕ = ⋂𝒫ⁱⁿᵈ(Z)
P(0) ∧ ∀ᴺk: P(k)⇒P(k+1) ⇒ ∀ᴺn: P(n)
Proving P(0) ∧ ∀ᴺk: P(k)⇒P(k+1) for ⋂𝒫ⁱⁿᵈ(Z)
and then concluding ∀ᴺn: P(n) for ⋂𝒫ⁱⁿᵈ(Z)
IS _a proof by induction_
A proof.by.induction conclusion claim ∀ᴺn:P(n)
only for each _element_ n of ℕ = ⋂𝒫ⁱⁿᵈ(Z)
It is silent with respect to P(ℕ)

The other is my proof by induction:
Assume that
the set F of FISONs F(n) = {1, 2, 3, ..., n}
has the union UF = ℕ.
Notice that
F(1) can be omitted
without changing the result.
Notice that
when F(k) can be omitted,
then also F(k+1) can be omitted.
This makes
the set of FISONs which can be omitted
without changing the result
an inductive set.
It has no last element.
It is F.
The complementary
set of FISONs which cannot be omitted,
has no first element.
It is empty.

From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.

No,
we have not obtained that.
https://en.wikipedia.org/wiki/Modal_fallacy
We have obtained that,
for each FISON F′ ∈ {F} and 𝒜 ∈ 𝒫{F}
if ⋃𝒜 = ℕ then _not.necessarily_ F′ ∈ 𝒜
{F} set of FISONs
𝒫{F} set of sets of FISONs
𝒫ᵁᐧᙿᴺ{F} set of ℕ.covering sets of FISONs
⋂𝒫ᵁᐧᙿᴺ{F} set of FISONs necessary to cover ℕ

From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.

 From the assumption ⋃{F} = ℕ
we have obtained ⋂𝒫ᵁᐧᙿᴺ{F} = {}
and also 𝒫ᵁᐧᙿᴺ{F} ≠ {}
𝒫ᵁᐧᙿᴺ{F} ≠ {}
𝒜 ∈ 𝒫ᵁᐧᙿᴺ{F}
⋃𝒜 = ℕ
ℕ ≠ {}
Set theory lives another day.

From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.
This result is false.

We have not obtained that result.
https://en.wikipedia.org/wiki/Modal_fallacy