Re: The set of necessary FISONs

On 2/10/2025 10:00 AM, WM wrote:

On 10.02.2025 13:37, Richard Damon wrote:

On 2/10/25 4:56 AM, WM wrote:

The set of useless FISONs is inductive and
therefore infinite.

Yes.
A FISON with FISONs.after is uselessᵂᴹ.
Each FISON is uselessᵂᴹ.
The set of FISONs is minimal.inductive.
The set of uselessᵂᴹ FISONs is minimal.inductive.

No FISON can change the assumption U(A(n)) = ℕ.

Yes.
For each FISON,
each FISON.number is in a FISON.after.
∀ᴺj′:∀ᴺi′:∃ᴺk′:
  k′ = max{i′,j′+1}
For each FISON,
the union of FISONs.after is the same set,
a set we have named ℕ.

Therefore
every FISON can be omitted.
==> { } = ℕ.

No.
Only  ⋃{FISONs.after} = ⋃{}
for a FISON without FISONs.after.
However,
each FISON is with FISONs.after.

Which means that each element is not needed,
but doesn't prove that you can't get the answer from
a union of an infinite set of them.

>
Does Zermelo define a set
by induction or
only its elements?

Zermelo defines exactly (not fuller, not emptier)
which elements are in a set,
a set of which there can only be one
(extensionality).
That defined set is minimal.inductive.
Therefore, induction is valid with it.