Re: The set of necessary FISONs

On 2/3/25 10:11 AM, WM wrote:

On 02.02.2025 21:11, Richard Damon wrote:
 

Since I don't recall you actually DEFINING what A(n) means, any assumption about it would be unwarranted.

 German A(n), or English F(n) is the FISON {1, 2, 3, ..., n}.
The assumption is that U(F(n)) = ℕ.

Then why did you change your notation.
You are just using the typical techiques of scammers and charlatons.
And if U is supposed to be the Union, that statement is just incorrect, there is no SINGLE FISON (your F(n) ) that makes the set of the Natural Numbers. After all, the input to the union operator is a collection of sets, not just a single set which is what F(n) would be.
What you need to do is take the union of a lot of set, in fact, and infinite number of them.
So, U(F(n), F(n+1), F(n+2), ...) would be N, not just U(F(n))

 By induction we prove that every F(n) can be removed without changing the union. Therefore the assumption leads to { } = ℕ. Therefore the assumption is wrong.

No, you prove that ANY FISON can be removed, not that ALL can be.

 Regards, WM

All you are doing is proving your math is just sloppy, and broken,