Re: The set of necessary FISONs

Am Tue, 11 Feb 2025 20:23:28 +0100 schrieb WM:

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.

Actually, no: you can not remove the whole set, only single elements.

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.

The erroneous step is from „every finite number” to „an infinite number”.

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.

Lol.

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?

It is wrong, which is the other possibility for a nonsensical consequent.

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.

No, it only covers the elements.

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

What is wrong in my application in your opinion?

You incorrectly include infinity.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.