Re: The set of necessary FISONs

On 26.01.2025 23:51, Richard Damon wrote:

Thinking a bit about this, the "Set of Necessary FISONs" will be empty,

Hence we can go through the sequence of FISONs one by one and eliminate each one so that none remains.

becuase no particular FISON is needed, you just need an infinite set of them.
 A simple proof of this is that we can build up the set at least two different ways using two infinite sets with no members in common.
 The first set it the set of all ODD FISONs, i.e. FISONs whoes highest member is an odd number.

We can go through the sequence of odd FISONs one by one and eliminate each one so that none remains.

 There is no Natural Number not covered by this union, as for every Natural Number, either it is itself odd, and thus part of its own FISON, or the number one greater than itself (which does exist) will be odd, and this number will exist in that FISON, and thus in the union of them.
 We can also do that with the set of even FISONs.

We can go through the sequence of even FISONs one by one and eliminate each one so that none remains.

For a FISON to be in the set of "Necessary" it would need to be in EVERY set that meets the requriment, but since no set does, there are no "necessary" FISONs.

And there are no sufficient FISONs. Each one can be discarded as insufficient.

 Just like there are no "necessaery" Numbers to have the sum of the set to be zero.

No. FISONs are ordered such that if F(n) is proven insufficient, we know that all smaller FISONs are proven insufficient too. Every FISON is insufficient, because ℵ₀ numbers are missing.
Regards, WM