On 11/25/2021 3:10 AM, Julio Di Egidio wrote:
> On 23/11/2021 01:10, Julio Di Egidio wrote:
> [Re: The non-existence of the set of all things that are NOT purple]
> > On Monday, 22 November 2021 at 22:51:52 UTC+1, Julio Di Egidio wrote:
> >> On Monday, 22 November 2021 at 21:04:13 UTC+1, Fritz Feldhase wrote:
> >>> On Monday, November 22, 2021 at 6:44:16 PM UTC+1, Julio Di Egidio
> wrote:
> >>
> >>>> Maybe it's trivial, but I'd expect some proof of
> >>>> existence of a least element or something...
> >> <snipped>
> >>
> >>> Note that we are working in the context of ZFC without
> >>> the Axiom of Foundation.
> >>> Indeed! We may restate the AoI the following way now:
> >>> EA succSet(A).
> >>> Finally, [Halmos] defines w [omega], the set of
> >>> (all) natural numbers:
> >>> | Since the intersection of every (non-empty) family of
> >>> successor sets is a successor set itself (proof?),
> >>> the intersection of all the successor sets included in A
> >>> is a successor set w.
> >>
> >> I'm with you/him that far.
Call this Lemma 1.
| If ∀s ∈ A, Succ(s) and A ≠ {} then Succ(∩A)
> >>> The set w is a subset of every successor set. If, indeed, B is an
> >>> arbitrary successor set,
> >>> then so is A n B. Since A n B c A, the set A n B is one of the sets
> >>> that entered into the definition of w;
> >>> it follows that w c A n B, and, consequently, that w c B.
> >>
> >> OK if by 'c' you did *not* mean proper subset, otherwise
> >> I don't see why that should be (generally) true.
>
> I was being polite:
> to be clear, the point was that's simply false.
You're wrong.
I suspect that the reason that disagreement is
typically expressed more tentatively in math and logic
circles is less a matter being polite and more
a matter of having been wrong in the past -- or,
more significantly, a matter of _having realized_
they were wrong in the past.
----
Define the set of successor subsets of A
sccsbs(A) = { s ⊆ A | Succ(s) }
Let A be a successor set.
Define
ω[A] = Intrsct(sccsbs(A))
Because intersection, ∀s ∈ sccsbs(A), ω[A] ⊆ s
Let B be any successor set.
By Lemma 1, Succ(A∩B)
A∩B ⊆ A
A∩B ∈ sccsbs(A)
ω[A] ⊆ A∩B
Also, A∩B ⊆ B, so
ω[A] ⊆ A∩B ⊆ B
ω[A] ⊆ B
Lemma 2.
∀B, Succ(B) -> ω[A] ⊆ B
> >>> The minimality property so established uniquely characterizes w
> >>
> >> Does it? Nor, on the other hand, I can see how we would prove that
> >> there is not just one successor set in that entire universe.
Let B be a successor set.
By Lemma 1, Succ(ω[B])
By Lemma 2, ω[A] ⊆ ω[B]
By a similar argument, ω[B] ⊆ ω[A]
( ω[A] ⊆ ω[B] ∧ ω[B] ⊆ ω[A] ) -> ω[A] = ω[B]
Lemma 3.
∀A,B, Succ(A) & Succ(B) -> ω[A] = ω[B]
Dropping '[A]', ω is the unique intersection
of all the successor subsets of any successor set.
Changing the successor set doesn't change ω
> > OK, it cannot be unique, we can add "non-standard zeros" (with their
> > successors), i.e. elements that are not 0 or a successor thereof, to
> > any successor set and it is still a successor set, QED. Then we might
Let A be a successor set.
Add non-standard zero {b} and its successors.
B = A ∪ { {b}, {b,{b}}, {b,{b},{b,{b}}}, ... }
Yes, B is a successor set.
No, that doesn't affect ω
Nonstandard {b}, {b,{b}}, ... are not in any
successor subset of A.
The intersection of successor subsets of B must
subset each successor subset of A.
No nonstandard {b}, {b,{b}}, ... is in any subset of A.
No nonstandard {b}, {b,{b}}, ... is in ω
> > any successor set and it is still a successor set, QED. Then we might
> > say the minimal successor set is the one that only contains "standard"
> > elements, those than can be reached from 0 by a finite number of
> > successor applications. -- I am still not convinced though, all that
> > talk of standard/non-standard: can we actually prove that an arbitrary
> > set X is *not* a "standard" element of successor sets?
>
> Enough thought: unless I am missing something, the answer is simply no.
> QED.
You are apparently missing the distinction between A
and Intrsct(sccsbs(A))
> For those who have missed the steps (kudos to FF for giving me the
> "basis"):
>
> 1. define successor of a set; // succ(x) = x union {x}
> 2. define generic successor set ("successor sets", SSs);
> // s is an SS iff 0 in s and, for all n, if n in s then succ(n) in s;
> 3. define standard vs non-standard elements of (any of) SSs;
> // 0 is standard and, for all x, if x is standard so is succ(x);
> // x is non-standard iff not x is standard;
There is conventionally an implicit minimality condition
in a definition like that. With a minimality condition,
standard and non-standard elements are exactly what
one would expect them tobe.
Without that convention, I could assert {b} is standard,
and you could not prove me wrong.
You could prove {b} was not in ω[B] = Intrsct(sccsbs(B))
but this does not contradict {b} being your "standard"
It seems to me that a minimality condition is called for,
implicit or otherwise. If you disagree, why do you disagree?