Re: The set of necessary FISONs

On 07.03.2025 21:11, Jim Burns wrote:

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.

I do not use indefinite followers but followers with n curly brackets. It can be read above.

How do you know which set?

 From that unique description:
{ } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.

 

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?

Yes. { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.

 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.

Is it so bewildering to replace "1" and "if n ∈ ℕ then n+1 ∈ ℕ" by Zermelo's notation?

 

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

>
Therefore there is no Bob.

>
Induction alone,

>
yields {} and its curly bracket followers.

 You have just now added another clause to
being inductive.

No.
Regards, WM