Re: The set of necessary FISONs

On 2/26/2025 6:55 AM, WM wrote:

On 25.02.2025 19:39, Jim Burns wrote:

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Zermelo for instance made such claims:
Um aber die Existenz "unendlicher" Mengen zu sichern,
bedürfen wir noch des folgenden, seinem wesentlichen Inhalte
von Herrn Dedekind herrührenden Axioms.
... Der Bereich enthält mindestens eine Menge Z, welche die Nullmenge als Element enthält und
so beschaffen ist, daß jedem ihrer Elemente a
ein weiteres Element der Form {a} entspricht
... Die Menge Z_0 enthält die Elemente 0, {0}, {{0}}, usw.
und möge als "Zahlenreihe" bezeichnet werden,
... Sie bildet das einfachste Beispiel einer
"abzählbar unendlichen" Menge.
[E. Zermelo: Untersuchungen über die Grundlagen der Mengenlehre I,
Mathematische Annalen (1908), S. 266]

⎛ But in order to ensure the existence of "infinite" sets,
⎜ we still need the following axiom,
⎜ the essential content of which comes from Mr. Dedekind.
⎜ ... The domain contains at least one set Z,
⎜ which contains the zero set as an element and
⎜ is such that each of its elements a
⎜ corresponds to another element of the form {a}
⎜ ... The set Z_0 contains the elements 0, {0}, {{0}}, etc.
⎜ and may be referred to as a "number series", ...
⎜ It forms the simplest example of a "countably infinite" set.
⎜ [E. Zermelo: Investigations on the Foundations of Set Theory I,
⎜ Mathematische Annalen (1908), p. 266]
⎝
-- google
The domain contains at least one set Z.
0∈Z ∧ ∀a:a∈Z⇒{a}∈Z
It is indefinite which set Z refers to,
apart from that claim definitely being true of Z.
That set Z has a subset Z₀ which might not be Z
(what is Z, after all?)
which may be referred to as a number series.
The number series Z₀ holds no extra numbers.
Suppose set W and Y are candidates for Z
0∈W ∧ ∀a:a∈W⇒{a}∈W
0∈Y ∧ ∀a:a∈Y⇒{a}∈Y
and
some elements are in W but not in Y.
Those elements in W\Y are extra, and not.in Z₀
Perhaps, outside the material you quote,
Zermelo has said that
Z₀ holds no extra numbers
or some version of that.
I haven't investigated what Zermelo has said.
Whatever he's said, the claims I've made here are true.
Because Z₀ holds no extra numbers,
Z₀ is definite, even though superset Z is indefinite.
That is, in my opinion, what makes all this work.
This next bit you (WM) might like, for a change.
It looks like the pseudo.induction.rule which
you have been trying to use.
If each set in a non.empty set.of.sets
is inductive,
then the intersection of the set.of.sets
is inductive.
⎛ ∀A∈𝒞≠0: 0∈A∧∀k:k∈A⇒k+1∈A  ⇒
⎝ 0∈⋂𝒞∧∀k:k∈⋂𝒞⇒k+1∈⋂𝒞
That is provable. Also,
that is the reason that Z₀ is definite,
even though Z is indefinite.
For inductive set W,
W₀ is the intersection of inductive.subsets of W
For inductive sets W and Y,
W∩Y ⊆ W ∧ W∩Y ⊆ Y is inductive.
W₀ = (W∩Y)₀ = Y₀
W₀ = Y₀ = Z₀ is unique, for all inductive sets.
Z₀ being the unique intersection of inductive subsets
makes
Z₀ its.own.only.inductive.subset,
which makes
Z₀ the only correct set to use in a proof by induction.