Is there any first order formula true in only one of R and R*?
I would think that if the answer is NO then R < R*.
What I'm exploring is the connection of < to [=], with the statement
that < implies [=].

Are there any other comparitive relations besides elementary embedding
that would fit with what I'm trying to do?  What I'm trying to do is
one major "leg" of my paper: there is a "superstructure" to all
structures.  What super means could be any comparitive relation.  But
what relation is 'good'?

On Dec 9, 8:12 am, Bruno Marchal <[email protected]> wrote:
> On 09 Dec 2010, at 05:12, Brian Tenneson wrote:
>
>
>
> > On Dec 5, 12:02 pm, Bruno Marchal <[email protected]> wrote:
> >> On 04 Dec 2010, at 18:50, Brian Tenneson wrote:
>
> >> That means that R (standard model of the first order theory of the
> >> reals + archimedian axiom, without the term "natural number") is not
> >> elementary embeddable in R*, given that such an embedding has to
> >> preserve all first order formula (purely first order formula, and so
> >> without notion like "natural number").
>
> > I'm a bit confused.  Is R < R* or not?  I thought there was a fairly
> > natural way to elementarily embed R in R*.
>
> I would say that NOT(R < R*).
>
> *You* gave me the counter example. The archimedian axiom. You are  
> confusing (like me when I read your draft the first time) an  
> algebraical injective morphism with an elementary embedding. But  
> elementary embedding conserves the truth of all first order formula,  
> and then the archimedian axiom (without natural numbers) is true in R  
> but not in R*.
>
> Elementary embeddings are *terribly* conservator, quite unlike  
> algebraical monomorphism or categorical arrows, or Turing emulations.
>
> Bruno
>
>
>
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> http://iridia.ulb.ac.be/~marchal/

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