Re: The set of necessary FISONs

On 10.02.2025 18:26, Jim Burns wrote:

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.

That is not the only reason. Also a FISON with no FISONs after it would be useless because
∀F(n) ∈ F: |ℕ \ F(n)| = ℵo.

 The set of FISONs is minimal.inductive.
The set of uselessᵂᴹ FISONs is minimal.inductive.

And there is no FISON beyond this set.

 For each FISON,
the union of FISONs.after is the same set,
a set we have named ℕ.

Then Zermelo does not describe the set ℕ but only a FISON. Then ℕ does follow upon it but, since it is not defined et all, does not exist.

 

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

 No.
Only  ⋃{FISONs.after} = ⋃{}
for a FISON without FISONs.after.

Name the first one. Every set of FISONs has a definable first element, unless it is empty.

 However,
each FISON is with FISONs.after.

Induction covers all.

 

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,

So do I.
 > a set of which there can only be one
 > (extensionality).
That is F.

 That defined set is minimal.inductive.
Therefore, induction is valid with it.
 

This set is infinite and has no successors after.
Regards, WM