Re: The set of necessary FISONs

On 26.01.2025 10:37, joes wrote:

Am Sun, 26 Jan 2025 09:41:22 +0100 schrieb WM:

On 25.01.2025 14:56, joes wrote:
>

There is obviously no single FISON that is equal to N.

Obviously every* union of FISONs is finite, because there are never two
consecutive actually infinite sets in ℕ, and every FISON has an infinite
set as successors.
Obviously every* union of FISONs is a FISON.

*finite
Every infinite union of FISONs is obviously infinite.

But no actually infinite set is possible.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality. It is bounded by the length of all rows all of which are finite. No actual infinity is represented by the first column.
Regards, WM

 

Do you agree that or every FISON the question whether it is necessary
can be answered?

Yes, in the negative. That does not imply the nonexistence of a
sufficient set.

It implies that there is not a first FISON. That proves by induction
that every FISON can be dropped. Your belief is dysfunctional logic.

Yes, every single FISON can be dropped.