Let $A$ and $B$ be models of the theory of Dense Linear Orders without Endpoints such that $|B| \subset |A|$. I'm trying to prove that $B$ is an elementary submodel of $A$. Using Tarski-Vaught test I was able to prove the following:
Proposition 1: If $A$ and $B$ are $L$-structures, $|B| \subset |A|$ and for every finite sequence $b_1,\ldots,b_n$ of elements of $B$ and an element $a$ of $|A|$ there is an automorphism (elementary embedding of $A$ onto itself) $\Phi$ of $A$ such that $\Phi(b_i) = b_i$ for $1 \leq i \leq n$ and $\Phi(a) \in B$ then $B$ is an elementary substructure of $A$.
I was trying to use Proposition 1 to my specific problem and I actually have a pretty good idea of what the automorphism should look like, but I can't construct it. Given $b_1,\ldots,b_n$ elements of $|B|$ and an element of $|A|$, if $b_i < a < b_j$ for some $i$, $j$, then I can associate $a$ with $b \in (b_i, b_j)$. And having constructed $\Phi(x)$ for a finite numbers of elements $x_1,\ldots, x_n$, I can construct $\Phi(x_{n+1})$ just respecting the order of $x_1,\ldots, x_{n+1}$. My problem is: $|A|$ can have any cardinality, but this procedure works only for countable universes.
Any ideas on how to fix my attempt?
Take $A = \mathbb{R} \times \mathbb{Q}$ ordered lexicographically and $B = \{0\}\times\mathbb{Q}$, Then the hypotheses of your proposition don't hold, e.g., with $n = 1$, $b_1 = (0, 0)$ and $a = (1, 0)$, there is no automorphism of $A$ that fixes $b_1$ and maps $a$ into $B$.