Re: The set of necessary FISONs

Am Tue, 11 Feb 2025 10:51:27 +0100 schrieb WM:

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.

If there were a last FISON, it would be N.

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

And there is no FISON beyond this set.

Nor are there natural numbers.

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.

Wrong. How did you even come up with this shit?

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.

Pick an endsegment of FISONs.

However, each FISON is with FISONs.after.

Induction covers all.

All finite sets of FISONs, sure.

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.

No, you try to include infinity.

 > 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.

A single set doesn’t have any successor at all.

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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.