Re: The set of necessary FISONs

On 3/7/2025 11:07 AM, WM wrote:

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.

<<JB<WM>>>

>
You need not the intersection however
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.

>
No.

<</JB<WM>>>

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.

I said (way back when) that
maybe you (WM) are sneaking ℕ in through the back door.
When you have described what you mean by
an indefinite curly.bracket.follower of {}
(which you haven't done yet),
you will find that you have described
an indefinite natural number.

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.

It is relevant where nothing more is known than
that a set is inductive.
You (WM) claim that induction alone 'makes' a set.
Which set, given only that it's inductive?
How do you know which set?

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.

Can you say what those are without intersection?
I suspect it can be said,
but I don't know how, right now,
and I'm sure that you (WM) haven't said how.

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

>
Therefore there is no Bob.

>
Induction alone,

>
yiels {} and its curly bracket followers.

You have just now added another clause to
being inductive.

And now try to find a difference to
F(1) and F(n) --> F(n+1).

FISONs are finite von Neumann ordinals.
We can describe the finite von Neumann ordinals,
and
we can prove,
without having a set of FISONs or of FISON.ends,
that a predicate inductive in the FISONs
is true without exception in the FISONs.
That result is what I'm calling 'induction'.
I think that's the best I can do until
I figure out what you're saying.