Re: The set of necessary FISONs

On 1/23/2025 3:43 AM, WM wrote:

On 22.01.2025 18:17, Jim Burns wrote:

On 1/22/2025 5:29 AM, WM wrote:

 [...]

>
For each (finite) FISON, there is
a (finite) FISON larger than it,
and
it is a proper subset of that larger FISON,
which is a proper subset of ℕ

>
Right.

ℕ is a superset of each FISON
Each set which is a superset of each FISON
  is a superset of ℕ
That's a longer.winded way to say that
ℕ is the union of all FISONs
Each FISON is a proper subset of ℕ
Each FISON is not ℕ
The claim
⎛ For each (finite) FISON, there is
⎝ a (finite) FISON larger than it.
is
prior to the claim
⎛ ℕ is the union of all FISONs
⎝ and isn't any FISON
and
that prior claim is not denied by
denying ℕ exists.

The sequence of FISONs is potentially infinite.
For each FISON F(n) there exists F(n^n^n).

KING BOB!!!
https://youtu.be/jK2XzKDab0E?si=ZROiFjKMTmiaGqPz&t=43
'Potentially infinite' forces into existence
the 'problem' you (WM) see.
Galileo Galilei saw the 'problem'
three hundred years before
Georg Cantor DIDN'T create the 'problem'.
⎛ 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.
Denying ℕ exists does not
deny that each FISON.end exists.
'Bye, Bob!

For the union of all (finite) FISONs,
there isn't any
(finite) FISON larger than it.

>
There is no constant "all" in potential infinity.

There is a finite.length description of a FISON.
Each object which it describes is a FISON.
Only objects which it describes are FISONs.
That is our 'all'.
It is sufficient to not.ignore the 'problem'.
('Bye, Bob!)
⎛ It is ordered such that each non.empty subset
⎜ holds a minimum and a maximum.
⎜ Except for ends, for each n, there is
⎝ n+1 first.after and n-1 last.before.
That description, at least, is visible,
whatever you (WM) mean by 'visible'.
We see it.

The union of all (finite) FISONs is not finite.

>
Anyhow
there is no set of FISONs
the union of which would be ℕ.

The 'problem' was seen before ℕ was mentioned.
It continues, even if ℕ is denied.

Each can be dropped.
All cannot be dropped.

>
"All" is not more than the repeated "each".

'All' is complete,
whatever you (WM) mean by 'complete'.
 From the finite.length description of a FISON,
we know that up to a FISON
is not complete, is not all.