Re: What is the 1st order formal system known as PA?


MoeBlee wrote:
> Rupert wrote:
> > MoeBlee wrote:
> > > Yes, I overlooked that. To speak of bounds on subsets of the domain, we
> > > have to speak of sets, so that we're in set theory or we can cast the
> > > matter in second-order. But, to express subsets of the domain couldn't
> > > we use an axiom schema with phi, for formulas, instead of using sets or
> > > quantifying over second-order predicate symbols; thus we could express
> > > the axioms of a complete ordered field as a first order theory with
> > > axioms and one axiom schema?
> > >
> >
> > I think, in fact, this theory you're talking about is the same as the
> > theory of real-closed fields. These are the ordered fields in which
> > every positive element has a square root and every polynomial of odd
> > degree has a root. The theory of real-closed fields is actually
> > complete.
>
> Is that the same theory as just taking the axioms for an ordered field
> and adding the least upper bound principle?
>

You were proposing to express the least upper bound principle by a
first-order axiom schema. If we do that, then, yes, as I said I think
the resulting theory is the same as the theory of real-closed fields.

> > There is no contradiction with Goedel because the notion "is
> > a natural number" cannot be defined in the first-order language of the
> > theory of real-closed fields, so arithmetic cannot be embedded in the
> > theory.
>
> That I understand. But I wasn't referring to incompleteness.
>
> Anyway, I know that the Los-Vaught test is that If T has only infinite
> models and is categorical in at least one cardinality, then T is
> complete. But what about the other direction? If T is complete, then
> what do we know about categoricalness?
>

I don't know. I doubt that the converse of your result holds.

> > And it is rather trivial to prove there exist nonisomorphic
> > real-closed fields.
>
> You mean that there are non-isomorphic models of the axioms for
> complete ordered fields (or real-closed fields; are these different?)?

What do you mean by the axioms for a complete ordered field?

A real-closed field is different to a complete ordered field.

You were proposing to replace the second-order version of the least
upper bound principle by a first-order axiom schema. If we do this, the
resulting theory is considerably weaker and not every model is a
complete ordered field. I wouldn't call the axioms you were talking
about "the axioms for complete ordered fields".

Yes, there are nonisomorphic models of the axioms for real-closed
fields.

> We know this from Lowenheim-Skolem, right? But what I am failing to
> picture is that there are non-isomorphic models of the axioms, yet in
> set theory we show that any two complete ordered fields are isomorphic.
>

All models of the second-order theory of complete ordered fields are
isomorphic.

There isn't really such a thing as a "first-order theory of complete
ordered fields". But you mentioned a first-order theory, namely the
first-order theory obtained by weakening the second-order form of the
least upper bound principle to a first-order axiom schema. There are
nonisomorphic models of this theory by Loewenheim-Skolem, yes. But not
all of them are complete ordered fields (because the first-order axiom
schema is not as strong as the second-order axiom). As I say, I think
this theory is the same as the theory of real-closed fields, so the
models would be precisely the real-closed fields.

> Thanks,
>
> MoeBlee

.

Relevant Pages

  • Re: What is the 1st order formal system known as PA?
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  • Re: What is the 1st order formal system known as PA?
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