Re: The set of necessary FISONs

Am Mon, 03 Mar 2025 09:57:47 +0100 schrieb WM:

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).

No. Z_0 is equivalent to N.

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.

sqrt(2) has an infinite decimal representation, like all reals. It just
doesn't have the period (0).

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.

And in the same way that Z_0 doesn't contain infinite elements, the set
of removable sets of FISONs doesn't.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.