Suppose $\frak A \prec \frak B$ are models in a language $\mathcal L$, and $|\frak A| < \kappa < |\mathcal L| \leq |\frak B|$. Is there an elementary chain $\frak A \prec \frak C \prec \frak B$ such that $|\frak C| = \kappa$?
Strengthen Lowenheim-Skolem
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$\begingroup$
logic
set-theory
model-theory
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2Not sure why this was downvoted . . . – 2017-02-23
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0As a rule of thumb, model theory has very little to say about models that are smaller than the size of the language. Of course, you can understand specific examples (though even to do this, you might need some set theoretic hypothesis, as suggested by Noah's answer!), but there are almost no general theorems. – 2017-02-24
1 Answers
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It is at least consistent that the answer is "no". This answer of mine gives an example of a theory in a continuum-sized language with a countable model $\mathfrak{A}$ such that any proper elementary extension is of size at least continuum. If $2^{\aleph_0}>\aleph_1$, then taking $\kappa=\aleph_1$ gives a counterexample to the question.
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3Also, if $2^{|A|} = |A|^{+}$, then we can assume that $|L| \leq |A|^{+}$ so that there are at most $|A|^{+}$ Skolem functions to close $A$ under. It follows that some violation of GCH is needed to produce a counterexample. – 2017-02-24