DLO without endpoints is model complete

Question

Let $T$ be the theory of dense linear orders without endpoints in the language $\{ < \}$. I need to show that if $\mathcal{A}$, $\mathcal{B}$ are two models of $T$ with $\mathcal{A} \leqslant \mathcal{B}$ then $\mathcal{A} \preccurlyeq \mathcal{B}$. I can do this by showing $T$ has quantifier elimination but the question suggests using the downward Lowenheim-Skolem theorem.

In the proof of the downward LS theorem, we take a subset of a structure and close it under Skolem functions, and then use this as the domain of the elementary substructure. But if we could show $\mathcal{A}$ is closed under Skolem functions then we'd have that it's an elementary substructure already by the Tarski-Vaught criterion.

How else can this be done?

Answer 1

We use Lowenheim-Skolem to reduce the general case to the countable case. Specifically:

If there are $\mathcal{A,B}\models DLOWOE$ with $\mathcal{A}\le\mathcal{B}$ but $\mathcal{A}\not\preccurlyeq\mathcal{B}$ then there are countable such $\mathcal{A},\mathcal{B}$.

Proof idea: Fix a tuple $\overline{a}\in\mathcal{A}$ witnessing the failure of elementarity and consider a countable elementary substructure of the expansion of $\mathcal{B}$ by a predicate naming $\mathcal{A}$ and constants naming $\overline{a}$.

Note that this is true for any theory - nothing special about $DLOWOE$ is being used here.

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We now handle the countable case directly, via homogeneity:

If $\mathcal{A}\le\mathcal{B}$ are countable DLOWOEs, then for every $\overline{a},b$ in $\mathcal{B}$ with $b\not\in\overline{a}$ there is some automorphism $\eta\in Aut(\mathcal{B})$ which fixes $\overline{a}$ pointwise and has $\eta(b)\in\mathcal{A}$.

Proof idea: Back-and-forth.

This lets us apply Tarski-Vaught: if $\overline{a}\in\mathcal{A}$ such that $\mathcal{B}\models\exists y(\varphi(\overline{a}, y))$, pick some $b\in\mathcal{B}$ with $\mathcal{B}\models\varphi(\overline{a},b)$ and apply the result above.