Re: The set of necessary FISONs

On 11.02.2025 18:42, Jim Burns wrote:

On 2/11/2025 4:31 AM, WM wrote:

On 10.02.2025 16:16, Jim Burns wrote:

The set F of FISONs which can be removed without
changing the assumed result UF = ℕ
is the infinite set F of all FISONs.

 Yes.

Fine.

This is proven by just the same induction
as Zermelo proves his infinite set Z.
>
Either you accept both proofs or none.
But without there is no set theory.

 That part above is fine.
Your problem is in the step after that,
which, for some reason, you skip over.

The part above defines the set of all FISONs that can be omitted without changing the union. There is no further step.

 Only a hypothetical last.FISON ͚F,
a FISON with {after.͚F} = {}
supports your reasoning:

The set of all FISONs is accepted. No further restricitons allowed.

There is no last FISON.
⋃{} ≠ ℕ

That is not claimed. Claimed is only UF = ℕ ==> ⋃{} = ℕ.
Can't you understand the meaning of an implication?

No, I won't try to dive into your private notation.

 Do you accept that,
for each two FISON.numbers j′ and i′
there exists a FISON.number maximum k′ of
i′ and the successor j′+1 of j′
?

It is irrelevant what details exist. Induction covers the whole infinite set.

How induction works is not well known to you (WM).

What is wrong in my application in your opinion?
Regards, WM