Re: The set of necessary FISONs

On 02.03.2025 21:28, Jim Burns wrote:

On 3/2/2025 1:52 PM, WM wrote:

On 02.03.2025 18:32, Jim Burns wrote:

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

 

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.

>
It has, namely ω, ω/2, etc.

 ω is outside Z₀
ω cannot be a counter.example to
a claim about an indefinite element of Z₀

You are right. I argued above concerning Cantor's actually infinite ℕ. It has the undefinable elements ω/2, ω/10, ω/20 inside and ω outside.
Z₀ does not contain undefinable elements. It contains simply all numbers that have FISONs (like v. Neumann constructs the FISONs directly).

 ω/2 is outside Z₀ and outside the ordinals.
ω/2 cannot be a counter.example to
a claim about an indefinite element of Z₀, or to
a claim about an indefinite ordinal.

Right. All in Z₀ is definable.
∀n ∈ Z₀: |ℕ \ {1, 2, 3, ..., n}| = ℵo

 

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.

>
Yes.
But it has no decimal representation.

 √2 splits all finite decimal representations.
Each is < √2 or > √2

Yes.

Therefore,
two such decimal.splitting.points √2 and √2′
don't exist.

Yes, but the topic is this: In exactly the same way as Z₀ is constructed by its elements, the set of removable FISONs is constructed by its elements, namely by induction.
Regards, WM