Re: The set of necessary FISONs

On 20.02.2025 20:05, Jim Burns wrote:

On 2/20/2025 5:32 AM, WM wrote:

On 20.02.2025 03:23, Richard Damon wrote:

 

>

>
Assume a set of sufficient FISONs.

 == Assume S ⊆ {F} exists such that
⋃S is the only.inductive.subset of ⋃S
 

|ℕ \ {1, 2, 3, ..., n}| = ℵo
is true for all FISONs.

 Yes.
 ⎛ For any two FISONs {i:i≤j} and {i:i≤k},
⎜ their sum {i:i≤j+k} is a FISON.

Yes.

That contradicts the assumption.

 What is the assumption?

The assumption is the existence of S.

 What is the contradiction?

The contradiction is that induction proves every FISON useless and therefore S  not existing.
Regards, WM