Re: The set of necessary FISONs

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

On 21.01.2025 21:35, Jim Burns wrote:

On 1/21/2025 6:45 AM, WM wrote:

[...]

>
For each finite set, there is
a (finite) FISON larger than it,
and that FISON subsets ℕ

For each (finite) FISON, there is
a (finite) FISON larger than it,
and that larger FISON subsets ℕ

It is a proper subset

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

and therefore not necessary
but completely useless in the union.

For each FISON,
each natural number in it
is also in a later, fuller.by.one, and larger FISON,
and
is not dropped by dropping that (earlier) FISON.

For each finite set, there is
a (finite) FISON larger than it.

>
That means the set is potentially infinite.

For only each finite set, there is
a (finite) FISON larger than it.
For only each finite set,
emptier.by.one sets are smaller and
fuller.by.one sets are larger.
For the union of all (finite) FISONs,
there isn't any
(finite) FISON larger than it.
The union of all (finite) FISONs is not finite.
For the union of all (FINITE) FISONs,
emptier.by.one sets are NOT smaller and
fuller.by.one sets are NOT larger.
For two sets which no FISON is larger than and
which differ only by Bob in one and not the other,
they are NOT sets such that
emptier.by.one sets are smaller and
fuller.by.one sets are larger.
Therefore,
each larger.than.any.finite  Bob.different set
is not smaller or larger than
the other larger.than.any.finite Bob.different set.

All can be dropped.

>
Each is not all.

>
All consists of each and not more.

Anything which is a finite ordinal
is in the set of all finite ordinals.
Anything which is not a finite ordinal
is not in the set of all finite ordinals.

Each can be dropped.
All cannot be dropped.

>
If you were right,
there would be a first FISON required
according to Cantor's theorem.

According to Cantor's theorem,
the opposite of your claim.
Each can be dropped.
The set {required.FISON} of all required FISONs F
  such that U{FISON} ≠ ⋃({FISON}\{F})
is empty.
There is no first FISON in {required.FISON} = {}
The set {FISON} of FISONs is not a FISON.

But according to Cantor's Theorem B,
every non-empty set of different numbers of
the first and the second number class
has a smallest number, a minimum.
This proves that
the set of indices n of necessary F(n),
by not having a first element,
is empty.

>
ℕ = ⋃{FISON}

>
Contradicted by mathematics,
namely Cantor's theorem.