Re: The set of necessary FISONs

On 2/23/2025 5:34 AM, WM wrote:

On 22.02.2025 21:29, Jim Burns wrote:

On 2/22/2025 5:17 AM, WM wrote:

We can remove all numbers from Z_0
and produce the empty set.

>
We can remove all numbers from {137}
and produce the empty set.
>
{137} ≠ 137

>
If we remove all numbers from {137}
we produce { },
if we remove all numbers from 137
nothing remains.
It is a matter of taste
whether the empty set is nothing.

Nothing is what is in {}.
{{}} ≠ {}
Nothing is not what is in {{}}.
The set of all FISONs with
each omissible FISON omitted
is {F}\{F} = {}
The set of all FISONs with
the set of omissible FISONs omitted
is {F}\{{F}} = {F}
∀F′ ∈ {F}:  F′ ≠ {F}  ∧  F′ omissible
{F} not omissible
----

The usable SUBSET ℕⁱᵒᵒⁱˢˢ of Zermelo's asserted Z
is its.own.only.inductive.sub.set (iooiss).
Our sets do not change.
>
I'm pretty sure your ℕ_def = ℕⁱᵒᵒⁱˢˢ
Our sets do not change.

>
Yes, except that
Zermelo starts with zero or empty set.
Please stop your silly labeling.
Zermelo calls it Z_0.

I point out that ℕⁱᵒᵒⁱˢˢ
is its.own.only.inductive.sub.set (iooiss)
in a fashion even more useful than how
you point out FISONs are finite.
Until you (WM) "remember" ℕⁱᵒᵒⁱˢˢ is iooiss,
the labels are needed.
----

For each two meanings
'inductiveₓ' and 'inductiveᵥ',

>
That is not a significant difference.

That is a significant lack.of.difference.
For meanings 'inductiveₓ' and 'inductiveᵥ',
with corresponding firstₓ, firstᵥ, sucₓ(), sucᵥ()
there are corresponding ℕₓⁱᵒᵒⁱˢˢ and ℕᵥⁱᵒᵒⁱˢˢ
such that
proving that a subset is inductive  is
proving that the subset is the whole set.
For ℕₓⁱᵒᵒⁱˢˢ and ℕᵥⁱᵒᵒⁱˢˢ
there is the map f: ℕₓⁱᵒᵒⁱˢˢ → ℕᵥⁱᵒᵒⁱˢˢ
⎛ firstₓ  ᶠ↦  firstᵥ
⎜ if ℕₓ ∋ x  ᶠ↦  v ∈ ℕᵥ
⎝ then sucₓ(x)  ᶠ↦  sucᵥ(v)
f is a bijection.
f is one.to.one
because sucₓ() and sucᵥ() are one.to.one
f is onto ℕᵥⁱᵒᵒⁱˢˢ
because image f(ℕₓ) ⊆ ℕᵥ is inductiveᵥ
thus f(ℕₓ) = ℕᵥ

Homework: Show the same for the set of FISONs.

{{}} ≠  {}