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 <marc...@ulb.ac.be> wrote:
> On 09 Dec 2010, at 05:12, Brian Tenneson wrote:
> > On Dec 5, 12:02 pm, Bruno Marchal <marc...@ulb.ac.be> 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.
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