Re: The set of necessary FISONs

On 2/20/2025 3:24 PM, Jim Burns wrote:

On 2/20/2025 2:27 PM, WM wrote:

On 20.02.2025 20:05, Jim Burns wrote:

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

 ⎛ 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}
 Are you denying that {i:j+1≤i≤k+1} exists?

Better: {i:j+1≤i≤j+k+1}  etc.
{i:i≤k} ⊆ ⋃S  ⇒  {i:j+1≤i≤j+k+1} ⊆ ⋃S\{i:i≤j}
#{i:i≤k} < #{i:j+1≤i≤j+k+1}

Are you denying that {i:j+1≤i≤k+1} is
larger than {i:i≤k} ?
 ⎜ ⋃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.

>
What is the assumption?

>
The assumption is the existence of S.
>

What is the contradiction?

>
The contradiction is that
induction proves every FISON useless

 Each FISON being uselessᵂᴹ (not.last) is
a consequence of ⋃S
being the only.inductive.subset of ⋃S
And vice versa.
 What are the TWO statements which
contradict each other?
 

and therefore S  not existing.