Re: The set of necessary FISONs

On 24.01.2025 16:44, Jim Burns wrote:

On 1/24/2025 4:37 AM, WM wrote:

On 23.01.2025 16:18, Jim Burns wrote:

 The union of all FISONs covers UF(n)

Simply contradicted by:
∀n ∈ UF(n): |ℕ \ {1, 2, 3, ..., n}| = ℵo
Try to find a counter example. Fail.

 Each FISON is a proper subset of another FISON.
Each FISON is a proper subset of UF(n)
No FISON is UF(n)

That is potential infinity.

 Whatever contains each FISON  contains UF(n)

Alas this is not a set but a (potentially in-) finite changing collection.

 

Otherwise Cantor's theorem would require
the existence of a first necessary FISON.

Do you know Cantors theorem?
Do you accept Cantors theorem:
"Theorem B: Every embodiment of different numbers of the first and the second number class has a smallest number, a minimum."
Do you agree that or every FISON the question whether it is necessary can be answered?

>

Each FISON is a proper subset of ℕ
Each FISON is not ℕ

>
Therefore each FISON can be dropped from
the set of candidates.

 Candidates for what office? For UF(n) ?

Candidates for the set of FISONs which are necessary to make UF(n) = ℕ.

Up to every FISON
|ℕ \ {1, 2, 3, ..., n}| = ℵo.

 For any two FISONs {1,2,...,j} {1,2,...,k}
their sum {1,2,...,j,j+1,j+2,...,j+k} is a FISON

Therefore the union cannot be larger than a FISON. The infinite union is the infinite FISON. But there is no infinite FISON

⎛ Consider Bob such that,
⎜ before all FISON.end.swaps n⇄n+1
⎜ Bob is in the first FISON.end 0
⎜
⎜ If Bob is in FISON.end n
⎜ then
⎜ it is after n-1⇄n and before n⇄n+1
⎜
⎜ If it is after all FISON.end.swaps
⎜ then Bob is not.in any FISON.end,
⎜ even though
⎜ no FISON.end.swap takes Bob
⎝ anywhere else.
 

Swaps cannot eliminate Bob. He remains but i the darkness.
Regards, WM