Re: The set of necessary FISONs

On 2/6/25 3:59 AM, WM wrote:

On 06.02.2025 03:29, Richard Damon wrote:

On 2/5/25 2:18 PM, WM wrote:

On 05.02.2025 19:43, Jim Burns wrote:
>

It might just matter what a natural number, induction,
a FISON, and a union are.

>
The axiom of induction:∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
>
P(1): U(F(n) \ F(1)) = ℕ.
>
P(k): U(F(n) \ {F(1), F(2), ..., F(k)}) = ℕ
==>
P(k+1): U(F(n) \ {F(1), F(2), ..., F(k+1)}) = ℕ.
>

And thus you can claim that no F(n) is "necessary".

 My proof is this: IF U(F(n)) = ℕ, THEN U({ }) = { } = ℕ.

But that is meaning less.
No ONE FISON makes the set of Natural Numbers, you need the union of an infinite set of them,.

>
Doesn't mean you can't use a set of them to build the set of Natural Numbers.

 It means precisely that. The premise is wrong because { } = ℕ is wrong.

Right, b ecause you started with error, because you started with nonsense, because you don't undetstand what you are talking about,
What do you think you are meaning by U(F(n))?

 Regards, WM