Re: The set of necessary FISONs

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

On 23.01.2025 16:18, Jim Burns wrote:

ℕ is a superset of each FISON
Each set which is a superset of each FISON
 is a superset of ℕ

3 is the integer between 2 and 4

Wrong.

Both claims are true, for the same reason.
That is the widely.used name for what's described.
As well, both claims are lacking in deeper meaning,
for the same reason. It's only a name.
Say "3 isn't the integer between 2 and 4"
And then, count 1,2,γ,4,...,12,1γ,14,...
and agree to γ+γ=6, γ×γ=9, γ×γ×γ×γ×γ=24γ
Oh! You've renamed 3
Other than your having made noises that _sound_ like
something that could use academic freedom,
there is no 'there' there.
Oh! You've renamed ℕ
Other than your having made noises that _sound_ like
something that could use academic freedom,
there is no 'there' there.
It's your (WM's) 'supercalifragilisticexpialidocious'.
https://www.youtube.com/watch?v=uZNRzc3hWvE&t=80s
⎛ He traveled all around the world and everywhere he went
⎜ He'd use his word and all would say there goes a clever gent
⎜ When dukes or Maharajas pass the time of day with me
⎜ I say me special word and then they ask me out to tea (woo)
⎜
⎜ Supercalifragilisticexpialidocious
⎜ Even though the sound of it is something quite atrocious
⎜ If you say it loud enough you'll always sound precocious
⎝ Supercalifragilisticexpialidocious

That's a longer.winded way to say that
ℕ is the union of all FISONs

>
The union of all FISONs does not cover ℕ.

Supercalifragilisticexpialidocious.
The union of all FISONs covers UF(n)
Each FISON is a proper subset of another FISON.
Each FISON is a proper subset of UF(n)
No FISON is UF(n)
Whatever contains each FISON  contains UF(n)

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

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) ?
That office must be vacant.
Darkᵂᴹ or visibleᵂᴹ,
each FISON is a proper subset of another FISON.

Nothing remains.
>

"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.

UF(n) contains each FISON.
No emptier set contains each FISON.
No smaller set contains each FISON.

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
UF(n) contains each sum of two FISONs.
No emptier set contains each sum of two FISONs.
No smaller set contains each sum of two FISONs.
UF(n)\{1,2,..,j} contains each {j+1,j+2,...,j+k}
No emptier set contains each {j+1,j+2,...,j+k}
No smaller set contains each {j+1,j+2,...,j+k}
|{j+1,j+2,...,j+k}|  =  |{1,2,...,k}|
No smaller set contains each FISON.
UF(n)\{1,2,..,j} is not smaller than UF(n)
UF(n)\{1,2,..,j} is not larger than superset UF(n)
|UF(n)\{1,2,...,j}|  =  |UF(n)|

Since every FISON is
the union of all its predecessors
we get
F(n): |ℕ \ UF(n)| = ℵo.

|UF(n)\UF(n)|  =  0
∀{1,2,...j}  ≠  UF(n)
|UF(n)\{1,2,...,j}|  =  |UF(n)|

If you don't believe in
the union of all F(n),
find the first exception.

Any set larger than each FISON
is not smaller than UF(n)
No exceptions.
UF(n) is sufficient to throw out your "logic".
⎛ 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.