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From: Chris Menzel on 2 Jul 2010 18:58
On Fri, 2 Jul 2010 14:40:34 -0700 (PDT), MoeBlee <jazzmobe(a)hotmail.com>
said:
> On Jul 2, 12:12 am, Transfer Principle <lwal...(a)lausd.net> wrote:
>
>> at what point beyond which a theory differs from ZFC such that we
>> should no longer call the objects which satisfy them sets?
>
> I don't know. But we can adopt certain definitions, such as:
>
> x is a set <-> (x=0 or Eyz y in x in z))
>
> That will work as long as the theory defines '0' appropriately.

And surely extensionality is essential to our conception of set.

>> This question has come up in other threads as well. By this line of
>> argument, one could even point out that there are objects satisfying
>> _ZF_ that are different from those satisfying ZFC, namely those
>> without choice functions, are infinite yet Dedekind finite, and so
>> on.
>
> ZF does not prove there exists an infinite yet Dedekind finite set,
> right? Rather, it is undecidable in ZF whether there exists such a
> thing. Same with sets without a choice function, right?

Of course, so all that can be said is that ZF simply cannot prove that
there are no such sets. It is misleading to say that there *are* such
sets that "satisfy" ZF (whatever exactly "satisfy" is supposed to mean).

From: Transfer Principle on 2 Jul 2010 23:46
On Jul 2, 3:58 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote:
> On Fri, 2 Jul 2010 14:40:34 -0700 (PDT), MoeBlee <jazzm...(a)hotmail.com>
> said:
> > ZF does not prove there exists an infinite yet Dedekind finite set,
> > right? Rather, it is undecidable in ZF whether there exists such a
> > thing.  Same with sets without a choice function, right?
> Of course, so all that can be said is that ZF simply cannot prove that
> there are no such sets.  It is misleading to say that there *are* such
> sets that "satisfy" ZF (whatever exactly "satisfy" is supposed to mean).

I mean that there's a model of ZF with such sets. Of course,
such a model would be a model of ZF+~AC, not a model of ZFC.

Note that Hughes was the first to use the word "satisfy" in
this manner:

Hughes:
Even in a not-too-distant from standard set theory like NF, it
seems fairly evident to me that objects which could conceivably
satisfy NF are somewhat different from objects that would satisfy,
say, ZFC.

So what does Hughes mean for an object to "satisfy" a theory? I
interpreted it to mean that there exists a model of NF which
proves the existence of the object, but not one of ZFC, but
maybe Hughes had something else in mind.

Thus, a D-finite T-infinite set "satisfies" the axioms of ZF
in that it satisfies Extensionality (i.e., it's determined by
its elements), satisfies Powerset and Union (since it has a
powerset and union), and so on. But maybe there's a better
word than "satisfies" to indicate that an object adheres to
each of a list of axioms.

All I wanted to know is whether objects whose existence is
refuted by ZFC but can exist in other theories should still
be called "sets." Menzel gives some criteria, namely that it
should at least adhere to Extensionality, and that sets ought
to contain elements (except 0) and be elements of other sets.
From: herbzet on 3 Jul 2010 03:44


Transfer Principle wrote:
> Chris Menzel wrote:
> > MoeBlee said:
>
> > > ZF does not prove there exists an infinite yet Dedekind finite set,
> > > right? Rather, it is undecidable in ZF whether there exists such a
> > > thing. Same with sets without a choice function, right?
>
> > Of course, so all that can be said is that ZF simply cannot prove that
> > there are no such sets. It is misleading to say that there *are* such
> > sets that "satisfy" ZF (whatever exactly "satisfy" is supposed to mean).
>
> I mean that there's a model of ZF with such sets. Of course,
> such a model would be a model of ZF+~AC, not a model of ZFC.
>
> Note that Hughes was the first to use the word "satisfy" in
> this manner:

No, it was me.

[herbzet]
> Even in a not-too-distant from standard set theory like NF, it
> seems fairly evident to me that objects which could conceivably
> satisfy NF are somewhat different from objects that would satisfy,
> say, ZFC.
>
> So what does [herbzet] mean for an object to "satisfy" a theory? I
> interpreted it to mean that there exists a model of NF which
> proves the existence of the object,

That is, a model in which the object exists.

That said, I agree with your understanding of my use of "satisfy".

> but not one of ZFC,

I don't think there are objects that at one and the same time
can satisfy (make true) the axioms both of NF and ZF(C). I could
be wrong about that.

> but maybe [herbzet] had something else in mind.
>
> Thus, a D-finite T-infinite set "satisfies" the axioms of ZF
> in that it satisfies Extensionality (i.e., it's determined by
> its elements), satisfies Powerset and Union (since it has a
> powerset and union), and so on. But maybe there's a better
> word than "satisfies" to indicate that an object adheres to
> each of a list of axioms.

It seems like a good word to me.

> All I wanted to know is whether objects whose existence is
> refuted by ZFC but can exist in other theories should still
> be called "sets."

Good question, matter of convention I guess.

> Menzel gives some criteria, namely that it
> should at least adhere to Extensionality, and that sets ought
> to contain elements (except 0) and be elements of other sets.

--
hz
From: Chris Menzel on 3 Jul 2010 04:49
On Fri, 2 Jul 2010 20:46:04 -0700 (PDT), Transfer Principle
<lwalke3(a)lausd.net> said:
> On Jul 2, 3:58 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote:
>> On Fri, 2 Jul 2010 14:40:34 -0700 (PDT), MoeBlee <jazzm...(a)hotmail.com>
>> said:
>> > ZF does not prove there exists an infinite yet Dedekind finite set,
>> > right? Rather, it is undecidable in ZF whether there exists such a
>> > thing.  Same with sets without a choice function, right?
>> Of course, so all that can be said is that ZF simply cannot prove
>> that there are no such sets.  It is misleading to say that there
>> *are* such sets that "satisfy" ZF (whatever exactly "satisfy" is
>> supposed to mean).
>
> I mean that there's a model of ZF with such sets. Of course, such a
> model would be a model of ZF+~AC, not a model of ZFC.
>
> Note that Hughes was the first to use the word "satisfy" in
> this manner:
>
> Hughes:
> Even in a not-too-distant from standard set theory like NF, it seems
> fairly evident to me that objects which could conceivably satisfy NF
> are somewhat different from objects that would satisfy, say, ZFC.

Not sure if the etymology is relevant here, but ok.

> So what does Hughes mean for an object to "satisfy" a theory? I
> interpreted it to mean that there exists a model of NF which proves
> the existence of the object, but not one of ZFC, but maybe Hughes had
> something else in mind.

So maybe something like this: An object s satisfies a theory T in a
model M of T iff there is a formula A(x) such that M |= A[x/s] (that is,
M satisfies A(x) when s is assigned to 'x'). So suppose M is a model of
ZF. Then we can say that s in M is "not an object of ZFC" if, for some
formula A(x), ZFC |- ~ExA(x) but M |= A[x/s].

> Thus, a D-finite T-infinite set "satisfies" the axioms of ZF in that
> it satisfies Extensionality (i.e., it's determined by its elements),
> satisfies Powerset and Union (since it has a powerset and union), and
> so on.

Ok, but you've dropped reference to any model here. The fact that, M |=
"s is infinite but D-finite" for some model of ZF doesn't mean that
there are, in fact, any such sets (assuming a realist view of sets). By
the Löwenheim-Skolem theorem, we can construct models of ZF+"there is an
infinite, D-finite set" out of, say, the finite von Neumann ordinals --
none of which, of course, is D-finite.

> All I wanted to know is whether objects whose existence is refuted by
> ZFC but can exist in other theories should still be called "sets."

I think it's clearer and less misleading to talk about theories instead
of objects, for the reasons just noted. A better way to put the
question, I think, is to ask at what point we would say that a given
theory is no longer a theory of *sets*.

> Menzel gives some criteria, namely that it should at least adhere to
> Extensionality, and that sets ought to contain elements (except 0)
> and be elements of other sets.

Minimally. But when one starts constructing a more definite picture,
notably, the cumulative conception of sets, a lot more would seem to be
required, e.g., foundation.

From: Nam Nguyen on 3 Jul 2010 10:04
Chris Menzel wrote:

> A better way to put the
> question, I think, is to ask at what point we would say that a given
> theory is no longer a theory of *sets*.

Would you share with us what that threshold point might be? Thanks.
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