Re: The set of necessary FISONs

On 3/2/2025 4:48 AM, WM wrote:

On 01.03.2025 21:10, Jim Burns wrote:

On 3/1/2025 1:28 PM, WM wrote:

On 01.03.2025 18:47, Jim Burns wrote:

On 3/1/2025 7:51 AM, WM wrote:

Z_0 contains only 0, {0}, {{0}}, ...

>
Z₀ = {0,{0},{{0}},...} is
the only subset of Z₀ which
holds 0 and, for each a, holds {a}

>
Z₀ is defined by induction.

>
inductive(Z)
>
inductive(Z) :⇔ 0∈Z ∧ ∀a:a∈Z⇒{a}∈Z
>
Z₀ is the emptiest (<EZ>"einfachste"?)
inductive set.

>
The simplest example.

I think Zermelo might be using 'einfachste==simplest'
in the same way that I'm using 'emptiest==leerste':
  as a pointer to the unique extremum of
  all sets with property P -- where P isn't size.
Z₀ is the simplest, the emptiest, but not the smallest.
  (Z₀ has same.sized inductive supersets.)
Simplicity is tricky, where membership is not,
so I will continue say Z₀ is emptiest.
⎛ Simplicity can serve another purpose.
⎜ It provides a reason for discussing Z₀ at all.
⎜  Yes, yes, yes, Z₀ is all the things you say,
⎜  but why should anyone care?
⎝  It's the simplest.

inductive(W) ⇒ Z₀ ⊆ W
inductive(Z₀)
>
Z₀ is not constructed by supertask.

>
Inductive Z₀:
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.

I wish to draw a distinction between
two senses of the verb 'construct'
'construct[make]' and 'construct[know]'.
Only 'construct[make]' is used for roads and bridges.
This is the more.commonly.used sense by far.
Only 'construct[know]' is used for abstract entities
such as ℕ and ℝ and ℂ.
We construct[know] Z₀ by finite.length proof.
Writing down and checking a finite.length proof
is not a supertask.
We do not construct[make] Z₀
In addition to Z₀ being an abstract object
and thus no more construct[make]able than {},
construct[make]ing Z₀ would require a supertask,
and we are finite non.supertaskers.

Z₀ = ⋂𝒫ⁱⁿᵈ(Z)

>
Also true.

Our inductive conclusions are justifiedly universal
because they are in narrower Z₀
not because of any supertask.

>
Induction abbreviates a supertask.
If 1 then 2, if 2 then 3, and so on. But supertasks will never pass through the dark numbers.

A claim about an indefinite element of Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
cannot have a counter.example outside of Z₀
That is the source of a matheologian's certainty.

They can only extend the defined numbers without end,
never crossing the infinitely larger domain of dark numbers
- if such exist at all!

Even if dark numbers exist, √2 remains irrational.
It is a matheological certainty that
there is no k ∈ Z₀⁺ such that k⋅√2 ∈ Z₀⁺
Dark numbers, if they exist, aren't in Z₀⁺