Re: The set of necessary FISONs

On 20.02.2025 21:24, Jim Burns wrote:

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

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

Induction covers all infinitely many elements.
Induction proves that no FISON is changing the assumed union ℕ.

 

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)

Useless is every FISON smaller than ℕ whether or not being last. (Note there is no last FISON.)

is
a consequence of ⋃S
being the only.inductive.subset of ⋃S
And vice versa.

Whatever

 What are the TWO statements which
contradict each other?

UF = ℕ and UF = ℕ ==> Ø = ℕ.
Regards, WM