Re: The set of necessary FISONs

On 28.02.2025 16:52, Jim Burns wrote:

On 2/28/2025 4:12 AM, WM wrote:

On 28.02.2025 01:00, Jim Burns wrote:

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

 

Zermelo's approach
does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)

>
It does.

 Eppur si muove.
 

Zermelo says it,

 Nope.

Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir noch des folgenden, seinem wesentlichen Inhalte von Herrn Dedekind herrührenden Axioms

and it is easy to prove it:
Adding all natural numbers established the set ℕ.

 We are finite beings. We do not do that.

Therefore we use FISONs without approaching ℕ.

How is Z accomplished?

 Z is NOT accomplishedᵂᴹ.

The existence of Z is secured by induction: Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir noch des folgenden, seinem wesentlichen Inhalte von Herrn Dedekind herrührenden Axioms.
Without induction Z is not existing.

 Zermelo describes the Z in the discussion

by induction. By what else?

{ } and a ==> {a}.

 True of Z because,
when Zermelo describes Z,
Zermelo describes such a set.

He describes it by induction.

 Z being such a set is not induction.

The proof of existence is done by induction.

 Induction proves inductive a subset of
a set which is its.own.only.inductive.subset.

The set Z and all its inductive subsets are proven by induction.

An inductive proof only proves about
a set which is its.own.only.inductive.subset,
like Z₀ and like ℕ, perhaps not like Z

Z contains many inductive subsets.

 

Proofs by induction are unreliable
in Robinson arithmetic.

>
Irrelevant.

 What you (WM) think is a proof by induction
is unreliable. But you don't care?

The set Z is not existing and not even defined without Zermelo's induction.
Note that for *definable* elements we have
{1, 2, 3, 4, 5} \ {1} \ {2} \ {3} \ {4} \ {5} = { },
which is same as
{1, 2, 3, 4, 5} \ {1, 2, 3, 4, 5} = { }.
Hence because of ∀n ∈ UA: |ℕ \ {1, 2, 3, ..., n}| = ℵo
we get
ℕ \ A(1) \ A(2) \ A(3) \ ... =/= { }
which is same as
ℕ \ UA =/= { }.
Regards, WM