Re: The set of necessary FISONs

On 08.02.2025 15:22, joes wrote:

Am Sat, 08 Feb 2025 11:27:49 +0100 schrieb WM:

On 08.02.2025 01:14, Richard Damon wrote:

On 2/7/25 11:39 AM, WM wrote:

On 07.02.2025 17:10, Jim Burns wrote:

On 2/6/2025 2:32 PM, WM wrote:

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I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).

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That's intended to be part of the definition of ℕ₁

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As well it is the definition of the collection of all FISONs.

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Which is curious, when one considers that the collection of all
FISONs appears nowhere in it.

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The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.

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So all elements are individually not needed.

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All not needed elements belong to an inductive set.

All FISONs do.

Yes, they all belong to the infinite set of FISONs which are not changing U(A(n)) = ℕ.

 

Therefore the set of FISONs which could have the union
ℕ has no first element.

WTF? The first element of set of all FISONs is A(0) = {0}.

Yes, the infinite set of useless FISONs is the set of all FISONs. But the first element of the set of FISONs which cannot be deleted is not existing.
Regards, WM