Re: The set of necessary FISONs

On 3/3/25 4:20 AM, WM wrote:

On 03.03.2025 01:42, Richard Damon wrote:

On 3/2/25 12:51 PM, WM wrote:

 

No, it means the set of REQUIRED FISONs is empty, not that we can't use a set of FISONs to make N.

 Every set of FISONs which is assumed to be ℕ must have a first element. No FISON can serve as such. That is proven by induction.

But the claim isn't that some FISON is N, it is that some infinite set of FISONs, none individually required, will union to N.
Since F(1) = {1} could be in that set, we can say the first element could be 1.
The problem is that none are individually required,

>

I did: UF = ℕ ==> Ø = ℕ.

>
But you didn't. You showed that there are no specific FISON requried to be in that set.

 I proved there is no FISON in that set that could be used with more reason than a cup of coffee. I did it by induction because then I can refer to Zermelo's construction of his potentially infinite sequence of numbers Z₀. Of course it follows also directly by
∀n ∈ Z₀: |ℕ \ {1, 2, 3, ..., n}| = ℵo
with Cantor's actually infinite ℕ.

No, you proved that none was a required element of the set, not that it coulddn't be used.
The problem is your trying to define a set with only required elements that unions to N doesn't need to exist, and can't be defined.
It doesn't need to exist, just like we can factor 36 using no "required" factors.
Your logic is just broken, and you are too stupid to understand the errors.

 Regards, WM