Re: how

Le 21/06/2024 à 02:18, Jim Burns a écrit :

On 6/19/2024 2:37 PM, WM wrote:

If the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
  does not hold a first element [in ⋂{Xᴬ⤾⁺¹₀}],
then the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
  is the empty set.

>
That is your big mistake!

 For subsets of the union ⋃{⟨0…n⟩} of FISONs
only the empty set does not hold a first element.
[1]

That is true.
Since from every FISON F(n) = {1, 2, 3, ..., n} we know by definition that it is neither sufficient nor necessary to make the union of FISONs ℕ, we can remove it from the union and find
∪{F(1), F(2), F(3), ...} = ℕ   ==>   ∪{ } = ℕ .
Among the FISONs of ℕ there is not, in any enumeration, a first one that is required to yield the union ℕ.
Usually this is apologized by the fact, that even in
{1, 2} ∪ {2, 3} ∪ {3, 1} = {1, 2, 3} (*)
it is impossible to find a first set which cannot be omitted from the union to yield {1, 2, 3}. But this argument fails. It is not a set of sets which is subject to Cantor's theorem B (every embodiment of different numbers of the first and the second number class has a smallest number) but only every set of ordinal numbers. Therefore we always have to enumerate the sets. In case of FISONs this is simple. We apply the natural order: {1, 2, 3, ..., n} --> n. Of course every other enumeration would also do. In case of the sets (*) we can use the written order from left to right. Then the first set not to be omitted is {2, 3} because after having omitted {1, 2} already, 2 would then be missing in the union. Every other order is possible and has a first set which cannot be omitted.
Regards, WM