Re: The set of necessary FISONs

On 2/19/2025 11:52 AM, WM wrote:

Am 18.02.2025 um 19:22 schrieb Jim Burns:

You (WM) have located a problem.
You try to work around it by not.mentioning it.
What you're not.mentioning is your assumption
that none of these sets are infinite.

>
Wrong.

Or one could read your posts.
<WM<RD<WM>>>

>
There is no element that could be
a meaningful member of any sufficient set.
Therefore there is no sufficient set.

>

Of course there are,
its just they are not individually needed,
but are collectively sufficient.

>
For every FISON there is the question:
Can it belong to a collectively sufficient set.
For every FISON the answer is no.
>

</WM<RD<WM>>>
Date: Tue, 18 Feb 2025 16:22:46 +0100
Message-ID: <vp28k5$1o28v$4@dont-email.me>
Yes.
There is no meaningful FISON in any sufficient set.
⎛ ¬∃F′ e {F}:
⎝ ∃S ⊆ {F}:  S ∋ F′  ∧  ⋃S = ℕ  ∧  ⋃Sᐠꟳᣟ ≠ ℕ
That implies and is implied by:
⎛ No FISON is ⋃{F} and
⎝ two FISONs are subset and superset, or vice versa.
You (WM) leap from that to
ᵂᴹ( There is no sufficient set.
That leap is where you apply your assumption
that there is no infinite set.

Induction has been invented for infinite sets.
Um aber die Existenz "unendlicher" Mengen zu sichern,
bedürfen wir noch des folgenden ... Axioms.
[Zermelo: Untersuchungen über
die Grundlagen der Mengenlehre I, S. 266]

What claims does Zermelo prove in the page or so
after he uses AXIOM VII "infinity"?

⋃{F} = ℕ

>
Proof:
If UF = ℕ is assumed, then
F(1) can be omitted
 without changing the union of the remainder.

⋃{F:F₁<F} = ⋃{F}

And if F(n) can be omitted
 without changing this union,
then also F(n+1) can be omitted
 without changing this union.

⋃{F:Fₙ<F} = ⋃{F}  ⇒  ⋃{F:Fₙ₊₁<F} = ⋃{F}

That makes the omitted

Better

That makes the [omissible] FISONs
the inductive collection of all FISONs

Yes.
{F′:⋃{F:F′<F}=⋃{F}} = {F′}

proves the implication:
If UF = ℕ, then { } = ℕ.

No.
Not proven for the set {F} of FISONs,
which is not a FISON.
The sum of any two natural numbers
is a natural number.
The "sum" of all the natural numbers,
for any reasonable definition of that,
will be larger than any natural number,
and not a natural number.