Re: The set of necessary FISONs

On 3/8/2025 3:45 AM, WM wrote:

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.

x ∈ {-1,0,1}
x is an indefinite reference.
We don't know x = -1
We don't know x = 0
We don't know x = 1
That isn't to say
that we know nothing about x
We know x³-x = 0, for example.
With further information,
either given or reasoned to,
an indefinite reference can become definite.
If we know only  x³-4x = 0
we know x ∈ {-2,0,2}
and that's indefinite,
but,
if we know both x³-x = 0 and x³-4x = 0
then we know x ∈ {-1,0,1}∩{-2,0,2} = {0}
and that's definite.
Indefinite.v.definite, the distinction between
an indefinite reference and a definite reference,
drives a lot of what I'm saying here.
Indefinite.v.definite
drives a lot of what set theory says.
x ∈ {-1,0,1} is an indefinite reference.
A = {-1,0,1} is a definite reference.
Zermelo's Axiom of Infinity describes Z
Z ∋ {} ∧ ∀a: Z ∋ a ⇒ Z ∋ {a}
Multiple sets satisfy that unique description.
That's an indefinite description.
The emptiest set satisfying Infinity describes Z₀
Only one set satisfies that unique description.
That's a definite description.
Indefinite.v.definite
is the reason that
'{i:A(i)} inductive subset of Z₀'
which is a definite description,
informs us about elements of Z₀
but
'{i:A(i)} inductive subset of Z'
which possibly is an indefinite description,
possibly doesn't inform us about elements of Z

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.

If only
{{{...{{{ }}}...}}} with finitely.many curly brackets
each with immediate.predecessor ⋃{{{...{{{ }}}...}}}
and with {} being one of its priors
are in Z₀
then, yes, that is a definite description.
And, yes, that seems to be what you meant.
If you ever want to make your descriptions clearer,
you won't be getting complaints from me.

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?

If n ranges over transfinite ordinals,
then your ℕ is unreliable in proof.by.induction.
It's not what Zermelo, von Neumann, or Cantor
mean by ℕ
If n ranges over only finite ordinals,
then
you didn't explicitly say that,
  which explicitness is the reason for writing,
and,
if you had explicitly said that,
you should have said somewhere
what a finite ordinal is,
and,
if you had said somewhere
what a finite ordinal is,
_where you said what a finite ordinal is_
is the place to look for a definition of ℕ
not the curly.bracket clauses,
which
makes what you actually wrote pointless.
But, other than that,
yeah, good job.