Re: The set of necessary FISONs

Am Fri, 31 Jan 2025 13:28:35 +0100 schrieb WM:

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.

???
Suppose I have a set whose union is N. Suppose further that there
is something smaller than all elements of that set. Then I can add
that element to the set, the union of that is still a superset of
N, and the additional element is not necessary.

A sufficient set does not imply a necessary subset;

If there is no first element necessary,

The empty set has no first element.

then all can be discarded,

No, only finitely consecutive ones.

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.

All naturals are finite. FISONs are not naturals but sets and can’t be
added together.

Removing all leaves nothing, in particular no sufficient set for
U(F(n)) = ℕ.

It is obvious that N is not empty.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.