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

Rupert wrote:
> > So any model of those axioms is a real closed field but not every model
> > of those axioms is a complete ordered field. Do I have that right?
>
> Yes.
>
> > So
> > it's a twist, no?, that it turns out that if you take the definition in
> > set theory of 'a complete ordered field' and restate it as first order
> > axioms, then its class of models is the class of real closed fields
> > instead of the class of complete ordered fields.
> >
>
> Well, it's not that surprising. The first-order axiom schema states
> that the least upper bound principle applies to those sets which are
> parametrically definable by formulas. There are only countable many
> such sets. So it's not surprising that this should turn out to be
> weaker than the full second-order least upper bound principle.

Right. But does that mean that all real closed fields are complete
ordered fields, but not vice versa? I'll address these fields in my
studies later. But for the purpose of this discussion about
isomorphisms and models, this would help to know.

> > I think I see this. All models are real closed fields, but there are
> > models in different cardinalities. Is that right?
>
> Yes.
>
> > And does it follow,
> > though, that all models of the same cardinality are isomorphic with one
> > another?
> >
>
> No, I don't think that's true. The algebraic real numbers are a
> denumerable real-closed field, but there are also non-archimedean
> denumerable real-closed fields.

Thanks. I'll check this out too eventually.

MoeBlee

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