Re: The set of necessary FISONs

Am Sun, 02 Mar 2025 18:51:05 +0100 schrieb WM:

On 02.03.2025 13:22, Richard Damon wrote:

On 3/1/25 1:09 PM, WM wrote:

On 01.03.2025 17:25, Richard Damon wrote:

On 3/1/25 9:31 AM, WM wrote:

On 28.02.2025 20:45, Richard Damon wrote:

On 2/28/25 11:44 AM, WM wrote:

>

No, Zermelo uses induction. I did not say that he uses the term
induction.

No, what you describe is NOT "induction".

Correct your qualifiers. Or look it up.
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite
inductive set F of FISONs.

You have a source that uses the term "Induction" for the recursive
iteration that builds the set?

∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
Wikipedia

Induction is the nucleus: P(0) /\ ∀k(P(k) ==> P(k+1)

 
And what SET did that build? P was a statement about a relationship.

P gives a relationship, for instance "is element of F". The universal
quantifier ∀ proves that all FISONs belong to the set that can be
removed.

Ah no, you're shifting the quantifier there. There are many sets of
FISONs that can be removed, one set per natural, containing all
FISONS of numbers less than that natural (also more sets of non-
contiguous FISONs). But those are all finite. The set of sets of
FISONs that can be removed together does not contain the set of
all FISONs (although it does contain the infinite sets of the odd
or even FISONs).

You can try to claim that the system it creates is inconsistent, but to
do so you need to show the error using the system as defined.

I did: UF = ℕ ==> Ø = ℕ.

--
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.