Re: The set of necessary FISONs

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.
⎜
⎜ For each {i:i≤k} ⊆ ⋃S, there is
⎜ a larger {i:j+1≤i≤k+1} ⊆ ⋃S\{i:i≤j}
⎜
⎜ ⋃S\{i:i≤j} is not.smaller than ⋃S
⎜
⎜ ⋃S\{i:i≤j} ⊆ ⋃S
⎜
⎝ ⋃S\{i:i≤j} is not.larger than ⋃S

That contradicts the assumption.

The subset {i′:#{i:i′<i}=#⋃S} of ⋃S is inductive.
⋃S is the only.inductive.subset of ⋃S
{i′:#{i:i′<i}=#⋃S} = ⋃S
∀k ∈ {i′:#{i:i′<i}=#⋃S}: #{i:k<i}=#⋃S
∀k ∈ ⋃S: #{i:k<i}=#⋃S

That contradicts the assumption.

What is the assumption?
What is the contradiction?