Re: The set of necessary FISONs

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

On 06.03.2025 20:03, Jim Burns wrote:

On 3/6/2025 12:05 PM, WM wrote:

On 06.03.2025 12:08, Jim Burns wrote:

On 3/6/2025 3:54 AM, WM wrote:

Am 06.03.2025 um 02:52 schrieb Jim Burns:

<<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>>>
>
No BECAUSE
that doesn't DEFINE Z₀

>
It defines Z₀ precisely.

>
It does not exclude Bob.
In order to exclude Bob,
some clause must state or imply Bob isn't in Z₀

>
No.
The first element is the empty set.
It does not contain Bob.
All following elements are
the empty set equipped with further curlybrackets.

You assume that
only
{} and its curly.bracket.followers are in Z₀
You didn't say that, above.
Instead, you said pretty much the opposite, above,
saying that intersection (making 'only') isn't needed.
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 {}
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.
⎜ For example,
⎜ {{Kevin}}, {{{Kevin}}}, {{{{Kevin}}}}, ...
⎜ the Zermelo.sequence of {{Kevin}}
⎜  is contained in
⎝ Kevin, {Kevin}, {{Kevin}}, ...
The description ∀a: a ↦ {a}  AND  e {}
says
that everything in the set
has its Zermelo.sequence in the set.
and
that there is AT LEAST the Zermelo.sequence of {}
That description is true of
{ only {}, {{}}, {{}}}, ...}
That description is true of
{ only {}, {{}}, {{}}}, ...  and
{ only Bob, {Bob}, {{Bob}}, ...}
That description is true of
{ only {}, {{}}, {{}}}, ...  and
{ only Kevin, {Kevin}, {{Kevin}}, ...}
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
After ONE OF those sets is asserted to exist,
a little more work shows that there is
a definite set with AT LEAST and AT MOST
{}, {{}}, {{{}}}, ...
That definite set is referred to as Z₀

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

>
Therefore there is no Bob.

Induction alone,
∀a: a ↦ {a}  AND  ∈ {}
doesn't say there is no Bob.
Induction alone says, for subset (i:A(i)}
  for a set we already know
  has no Bob and no Kevin and so on,
  that is, {only {}, {{}}, {{{}}}, ...}
that subset {i:A(i)} is
the whole  {{}, {{}}, {{{}}}, ...}

What does your inductive description permit in Z₀
which should not be permitted in Z₀ ?
Anything.
Your description permits Bob, as long as
{Bob}, {{Bob}}, {{{Bob}}}, ... are also in Z₀

>
Induction excludes Bob.

Induction alone does not exclude Bob.
'Emptiest inductive' excludes Bob.
With Bob and Kevin and so on already excluded,
induction (now alone) excludes all subsets but one.