Re: The set of necessary FISONs

On 1/31/25 7:28 AM, WM wrote:

On 31.01.2025 12:31, joes wrote:

Am Thu, 30 Jan 2025 23:32:52 +0100 schrieb WM:

On 30.01.2025 15:30, Richard Damon wrote:

On 1/30/25 4:14 AM, WM wrote:

On 29.01.2025 13:46, Richard Damon wrote:
>

We can in fact build an infinite set of infinite sets of FISONs whose
union is the set of Natural Numbers,

>

If there is a set with U(F(n)) = ℕ, then it has a first element that is
not completely useless.

Not necessarily.

 Necessarily, because otherwise all elements can be discarded.

Like all the factors of 36 can be discarded.

>

 A sufficient

set does not imply a necessary subset;

 If there is no first element necessary, then all can be discarded, and the rest is empty - not sufficient to yield U(F(n)) = ℕ.

And thus 36 can't be factordd.
Your logic is just incorrect.
There is no need for a "required set" to be able to build a set that does something.

But all F(n) can be shown to be completely
useless because infinitely many natnumbers are missing.

Again: every single one, or even an arbitrary finite number.
If you have inf. many segments, you obviously have inf. many
numbers.

 All finite natural numbers as well as all FISONs obey the Peano axioms. Removing all leaves nothing, in particular no sufficient set for U(F(n)) = ℕ.

But if your try to remove "all" the finite natural numbers, you are removing an infinite set, and the removal of an infinite set from an infinite set can leave nothing, or a finite set, or and infinite set, so your "contradiction" isn't one, but just shows your logic can't handle infinite sets.

 Regards, WM