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Could anyone of you explain me why $(\mathbb{R},<)$ is $\omega$-saturated?

EDIT: do you know also why the theory of Boole algebras without atoms is $\omega$-categoric?

Added: The added question about Boolean algebras is at Boole algebras without atoms

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en.wikipedia.org/wiki/Saturated_model –  user9413 May 16 '11 at 18:50
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I think you should ask your second question as a separate question. –  JDH May 16 '11 at 19:31
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up vote 12 down vote accepted

Your structure is an endless dense linear ordering, and this is a theory that admits elimination of quantifiers. Thus, every assertion in this language is equivalent to a quantifier-free assertion. If one has finitely many parameters, then the only possible consistent types are the assertions about how those parameters are ordered, and how the new variable $x$ fits into the resulting intervals. That is, the variable $x$ is either equal to one of them, or between two successive ones, or above all or below all of them. All such types are already realized in $\mathbb{R}$, since it has such points already for each of these possible patterns, and so the structure is $\omega$-saturated.

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A different approach: the theory of dense linear orderings is $\omega$-categorical.

It is a general (and easily seen) fact that if a theory is categorical in a given cardinality, the only model of this cardinality is saturated.

From that it is easy to see that if a theory is $\kappa$-categorical, then every model of the theory of cardinality at least $\kappa$ is $\kappa$-saturated (just extend the set of parameters to an elementary submodel of cardinality $\kappa$).

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