Re: The set of necessary FISONs

On 26.01.2025 18:55, joes wrote:

Am Sun, 26 Jan 2025 10:59:09 +0100 schrieb WM:

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.

The length? You gave an infinite list. Every column is infinite.

No row is infinite. No column is longer than all rows. All rows and columns belong to potentially infinite collections.

 

It is bounded by the length of all rows all of which are finite.
No actual infinity is represented by the first column.

The rows are unbounded.
 

Right. The columns are unbounded too. With n also n^n^n is contained. Nevertheless the fixed number ℵo of elements is neither in a row (by definition) nor in a columns (by symmetry).
Now try to find yourself the solution: What is unbounded but smaller than the first transfinite quantity?
Regards, WM
Regards, WM