Re: The set of necessary FISONs

Am Wed, 12 Feb 2025 11:19:55 +0100 schrieb WM:

On 12.02.2025 10:46, joes wrote:

Am Tue, 11 Feb 2025 20:23:28 +0100 schrieb WM:

On 11.02.2025 18:42, Jim Burns wrote:

On 2/11/2025 4:31 AM, WM wrote:

On 10.02.2025 16:16, Jim Burns wrote:

>

The set F of FISONs which can be removed without changing the
assumed result UF = ℕ is the infinite set F of all FISONs.

Actually, no: you can not remove the whole set, only single elements.

Induction proves all elements and even the set (Zermelo, Z).

No, N does not include itself.

The erroneous step is from „every finite number” to „an infinite
number”.

Induction proves an infinite number.

No, all naturals are finite!

Induction covers the whole infinite set.

No, it only covers the elements.

All elements. None remains.

Yes, there is no natural number of FISONs you can not remove.

Mithilfe der Definition über induktive
Mengen lässt sich die Beweismethode der vollständigen Induktion
rechtfertigen (daher auch der Name induktiv): Soll gezeigt werden, dass
alle natürlichen Zahlen eine bestimmte Eigenschaft haben, so betrachte
die Menge E ... [Wikipedia]

Says nothing about *the set* having that property.

How induction works is not well known to you (WM).

What is wrong in my application in your opinion?

You incorrectly include infinity.

No, there is no infinite FISON. All are finite.

Why do you want to remove an infinite FIS of FISONs then?

All can be omitted without changing the result.

No, any one can be, or even any finite number, but there are
infinitely many.

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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.