Re: The set of necessary FISONs

On 03.03.2025 12:54, joes wrote:

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

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.

How that? By induction we prove for every element n ∈ Z₀:
|ℕ \ {1, 2, 3, ..., n}| = ℵo

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

There is no actually infinite decimal representation. It would have ω digits. What digit would be at position ω/2?

 

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.

Of course not. Only finite FISONs are existing and removable. If all are removed, none remains. UF = ℕ  ==> Ø = ℕ.
Regards, WM