Re: The set of necessary FISONs

On 11.04.2025 04:03, Richard Damon wrote:

On 3/5/25 4:05 PM, WM wrote:

Therefore iteration fails to produce actual infinity.

 Where do you get that?

The strongest argument is the thinned out harmonic series. It is impossible to pass the dark numbers:
The harmonic series diverges. Kempner has shown in 1914 that when all terms containing the digit 9 are removed, the serie converges. Here is a simple derivation: https://www.hs-augsburg.de/~mueckenh/HI/ p. 15.
That means that the terms containing 9 diverge. Same is true when all terms containing 8 are removed. That means all terms containing 8 and 9 simultaneously diverge.
We can continue and remove all terms containing 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 in the denominator without changing this. That means that only the terms containing all these digits together constitute the diverging series.
But that's not the end! We can remove any number, like 2025, and the remaining series will converge. For proof use base 2026. This extends to every definable number. Therefore the diverging part of the harmonic series is constituted only by terms containing a digit sequence of all definable numbers.
The terms are tiny but that part of the series diverges. This is a proof of the huge set of undefinable or dark numbers.

The memeber of the set we iterate in are all finite, but the resultant set is infinite itself.

It is impossible to iterate the definable terms such that the remainedr is less than infinity. The never iterated Terms are almost all terms.
Regards, WM