# DLO without endpoints is model complete

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?

• The idea of choosing Skolem functions under which $\mathcal A$ is closed runs into the problem that you need Skolem functions for all formulas, of arbitrary quantifier complexity. I don't see how to handle those without first reducing the quantifier complexity, i.e., without first proving quantifier elimination. But that would defeat your purpose here. May 17, 2020 at 16:47

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.

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.

• DLOWE isn't really less ambiguous than DLO, is it? The W could stand for "with" or "without"! :0) May 17, 2020 at 20:28
• @AlexKruckman ... That's a real DLO woe right there. May 17, 2020 at 21:05