Re: The set of necessary FISONs

On 07.03.2025 16:08, Jim Burns wrote:

On 3/7/2025 4:23 AM, WM wrote:

You assume that
only
{} and its curly.bracket.followers are in Z₀

That is what the intersection of all Zermelo-inductive sets produces.

 You didn't say that, above.
Instead, you said pretty much the opposite, above,
saying that intersection (making 'only') isn't needed.

I said that the intersection isn't needed when instead of Zermelo's approach {} and its curly bracket followers are used.

  The description ∀a: a ↦ {a} of a set
says
that each thing in the set
has its Zermelo.sequence in the set.
 {}, {{}}, {{{}}}, ...
is
the Zermelo.sequence of {}

It is what Zermelo calls the sequence of numbers.

 Bob, {Bob}, {{Bob}}, ...
is the Zermelo.sequence of Bob.
 Kevin, {Kevin}, {{Kevin}}, ...
is the Zermelo.sequence of Kevin.
 ⎛ Everything in a Zermelo.sequence
⎜  has its Zermelo.sequence
⎜   contained in the over.sequence.

That is true but irrelevant when only {} and its curly bracket followers are used.

Zermelo's Axiom of Infinite asserts
the existence of ONE OF those or similar sets
with AT LEAST only {}, {{}}, {{{}}}, ...
That indefinite set is referred to as Z

And {} and its curly bracket followers is referred to as Z₀.

Induction says what's IN Z₀
More description is needed.

>
Therefore there is no Bob.

 Induction alone,

yiels {} and its curly bracket followers.
And now try to find a difference to F(1) and F(n) --> F(n+1).
Regards, WM
Regards, WM