Re: The set of necessary FISONs

On 2/27/2025 5:01 PM, WM wrote:

On 27.02.2025 21:41, Jim Burns wrote:

On 2/27/2025 2:50 PM, WM wrote:

On 27.02.2025 19:19, Jim Burns wrote:

This is what you (WM) have called induction:
⎛ Each inductive predicate A is true.without.exception.
⎜ ∀ᵖʳᵉᵈA: A(0) ∧ ∀k:A(k)⇒A(k+1) ⇒
⎝ ∀n:A(n)
 (1)

>
No, I call induction
a very restricted number of predicates.

>

I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).

>
Correct.

It incorrect that
_in each domain_
each inductive predicate is true.without.exception

But not necessary in its generality
for my purpose.

If one uses (1)
without regard for which domain one uses it in,
as you (WM) do,
then, necessary or not,
that is using it in its generality.

If A(n) is useless for UA = ℕ,
then A(n+1) us useless too.
No reason to extend this simple concept.

>
You extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
but you only claim it, you don't justify it.

>
I use Zermelo's approach
without wich there is no set theory.

Zermelo's approach (a standard approach)
is to assert
the existence of inductive Z by Infinity,
and of 𝒫(Z) = {S⊆Z} by PowerSet,
and of 𝒫ⁱⁿᵈ(Z) = {S⊆Z:inductive.S} by Separation,
and of ⋂𝒫ⁱⁿᵈ(Z) = {k∈Z:∀S∈𝒫ⁱⁿᵈ(Z):S∋k} by Separation.
Z₀ = ⋂𝒫ⁱⁿᵈ(Z) is its.own.only.inductive.subset.
Zermelo's approach does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)

By the same induction
I prove UF = ℕ  ==> Ø = ℕ.

>
What you use to prove that is
∀n:Aᴺ(n) ⇒ A(ℕ)

>
That is how Zermelo guarantees Z₀.

>
Zermello's Infinity guarantees a superset Z of Z₀

>
How is that accomplished?

By Zermelo's approach:
Z ⇒ 𝒫(Z) ⇒ 𝒫ⁱⁿᵈ(Z) ⇒ ⋂𝒫ⁱⁿᵈ(Z) = Z₀

From Z, it follows,
by Powerset and by Separation,
that Z₀ exists.
>
∀n:Aᴺ(n) ⇒ Aᴺ(ℕ)
is your fantasy.

>
It is Zermelo's approach.

It's not.even.wrong.
And it's not Zermelo's approach,
Z ⇒ 𝒫(Z) ⇒ 𝒫ⁱⁿᵈ(Z) ⇒ ⋂𝒫ⁱⁿᵈ(Z) = Z₀

You would find your posts greatly improved
by criticizing (if you can) _our_ reasoning,

>
You deny Zermelo's approach.

I deny your (WM's) not.even.wrong, non.Zermelo approach.

His Z is ensured by induction.

Nope.
Z ⇒ 𝒫(Z) ⇒ 𝒫ⁱⁿᵈ(Z) ⇒ ⋂𝒫ⁱⁿᵈ(Z) = Z₀

When we have shown that there is
the intersection of all inductive subsets of
an inductive set,
then we have constructed ℕ.

>
We don't even need the intersection
if we reduce Zermelo's approach to
Lorenzen's approach:
I is a natural number, and
if x is a natural numbers then x+1 is a natural number.

Consider Robinson arithmetic.
It's all that you say it should be.
Some models described by it
are NOT their.own.only.inductive.subset.
Proofs by induction are unreliable in Robinson arithmetic.
Wherever you got Lorenzen's approach from,
send it back.