Re: The set of necessary FISONs

On 1/25/2025 6:02 AM, WM wrote:

On 24.01.2025 16:44, 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.

UF(n) is
the union of all FISONs
The union of all FISONs covers
the union of all FISONs
Therefore, no.

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.

No.
A FISON is
  always and everywhere a FISON.
A set containing a FISON
  always and everywhere contains that FISON.
A set containing a set containing a FISON
  always and everywhere contains
  the set containing the FISON.
Therefore, no changes.
Anyway,
what you have to say about UF(n) doesn't matter
for the goal of preventing Bob from disappearing.
The FISONs themselves are reason enough
to throw out your "logic".
For each FISON ⟦0,n⟧
there is a fuller.by.one FISON ⟦0,n+1⟧  and
there is a swap n⇄n+1, ordered by n
For Bob who is in room 0 before all swaps,
if he is in room n
then it is between n-1⇄n and n⇄n+1
If it is after all swaps,
then Bob isn't in any FISON.end room,
even though swaps only place him in FISON.end rooms.

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

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

If Bob is somewhere after all swaps,
how did he get there?

Swaps cannot eliminate Bob.
He remains but i the darkness.

No swaps are into the darkness.