Re: The set of necessary FISONs

On 2/16/25 1:02 PM, WM wrote:

On 16.02.2025 18:23, joes wrote:

Am Sun, 16 Feb 2025 16:52:00 +0100 schrieb WM:

n.

What are you talking about? There are always inf. many if you remove
a natural number of FISONs, and their union is N.

 Induction proves that there is no FISON that, when removed, changes the premise. Therefore all FISONs can be removed without changing the premise. The conclusion is U(F\F) = ℕ.

Nope, jump from for any to for all, which isn't valid.

 

You DO change the result to the empty set if you remove everything; the
union of all FISONs is not the empty set.

 But the union of all FISONs which have the union ℕ is the empty set.

Nope,  you jumped conditions. You proved for any, not for all.

 

You can only remove a finite
number of them, doesn’t matter which.

 Wrong. Learn the meaning of induction.

Maybe you should do the same.
You need to be clear on what each step is actually doing, and what the resultant set actually is.

 Regards, WM