Re: The set of necessary FISONs

Am Mon, 03 Mar 2025 17:38:21 +0100 schrieb WM:

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

Like that.

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?

That is not a natural position.

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.

No, finite sets of them are removable (including sets containing only
a single FISON).

If all are removed, none remains. UF = ℕ  ==> Ø = ℕ.

That "all" you want to remove is not in the set.

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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.