Re: The set of necessary FISONs

On 2/12/2025 5:19 AM, WM wrote:

On 12.02.2025 10:46, joes wrote:

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:

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.

>
Induction proves all elements and
even the set (Zermelo, Z).

No.
Not the set.
Each member of the set.
Set intensionality means
same.membered sets are one set.
It does not mean sets are self.members.

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

>
Induction proves an infinite number.

Induction is valid without any infinite sets.
Consider ST+F,
George Boolos' tiny theory ST plus
an explicit ban F on infinite sets:
⎛ ST: {} and x∪{y} exist, and
⎜ same.membered sets are one set.
⎝ F: Emptier.by.one sets are smaller.
Induction on all the finite ordinals
is a theorem of ST+F.
It's wordier without infinite sets.
In ST+F, we can't say k is in ℕ, instead
we describe k in the language of set theory
as a finite ordinal, as a FISON.end.
Wordier, but still true.
In ST+F, it is a theorem that,
for predicate P(n),
if
any FISON.end counter.example n exists, ¬P(n)
then,
in the FISON ended by n,
a first counter.example 0 or k+1 exists.
∃ᴺn:¬P(n) ⇒ ¬P(0) ∨ ∃ᴺk:¬P(k+1)∧∀ᴺj≤k:P(j)
Formal.consequence and also.a.theorem:
P(0) ∧ ∀ᴺk:P(k)⇒P(k+1) ⇒ ∀ᴺn:P(n)
The induction theorem.
⎛ k+1 means k∪{k}
⎜ 0 means {}
⎜ ∃ᴺn:¬P(n) means
⎜ exists n such that ¬P(n) AND
⎜ either  n={}  or
⎜ n∋{} and,
⎜ for each non.{} j ∈ n∪{n},
⎝ exists i ∈ n such that i∪{i}=j
In ST+F, there are no infinite sets,
but there is induction.
Also, in ST+F, swaps exist such that,
after all those swaps,
Bob isn't anywhere Bob has swapped to.
⎛
⎜ ∀k⇄k+1 ∃k+1⇄k+2
⎜ For each swap.in, a later swap.out
⎜
⎜ Bob somewhere he has swapped to
⎝ is not after all those swaps.