Re: The set of necessary FISONs

On 1/30/2025 5:32 PM, WM wrote:

On 30.01.2025 15:30, Richard Damon wrote:

On 1/30/25 4:14 AM, WM wrote:

On 29.01.2025 13:46, Richard Damon wrote:

We can in fact build
an infinite set of infinite sets of FISONs whose union is the set of Natural Numbers,

>
Do it.
Build such a set.
I will show that it fails because
all FISONs are useless for reaching the aim.

>
I did.

This use of the term 'build' is
showing that an infinite road exists.
It isn't walking an infinite road.

If there is a set with U(F(n)) = ℕ,
then it has a first element that is not completely useless.

Only not.followed elements are not.completely.uselessᵂᴹ.
If there is a set with ⋃{F(n)} = ℕ,
then each element is followed,
and the subset of not.completely.uselessᵂᴹ elements is empty.

But all F(n) can be shown to be completely useless
because infinitely many natnumbers are missing.

None are missing from ⋃{F(n)}
⋃{F(n)} is the emptiest superset of each FISON.
There is no element of any FISON not.in ⋃{F(n)}
Each FISON is smaller than its fuller¹ FISON,
thus smaller than superset ⋃{F(n)} of its fuller¹.
⋃{F(n)} is larger than any FISON
The sum of any two FISONs is a FISON.
⋃{F(n)} is larger than the sum of any two FISONs.
Each end.segment ⋃{F(n)}\{1,...,j} is
larger than any FISON.
For each j in ⋃{F(n)}
all infinitely.many followers follow,
none are missing.

Therefore you did not.
You cannot.
Nobody can.

ℕ holds only completely.uselessᵂᴹ numbers.
ℕ isn't what you (WM) think it is.
That's not a logic.problem.
That's a you.problem.