# Prove that the class of dense linear orders cannot be axiomatized by purely universal sentenes [closed]

That is, prove that there is no set $$T$$ of purely universal sentences such that for every structure $$A$$ over the signature $$\{\leq\}$$, $$A$$ is a dense linear order iff $$A\models T$$.

## closed as off-topic by Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy PlamondonFeb 2 at 20:09

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• Given a model of such a theory, consider whether a substructure must be a model as well. – Jishin Noben Jan 26 at 9:36

Suppose there was such a set $$T$$. Clearly $$(\mathbb{Q}, \le) \models T$$, as it is a dense linear order.
$$(\mathbb{Z}, \le)$$ is a substructure of $$(\mathbb{Q}, \le)$$ so $$T$$ also holds in this, because purely universal sentences stay true on subsets: $$(\mathbb{Z}, \le) \models T$$.
But $$(\mathbb{Z}, \le)$$ is not a dense linear order, which contradicts the assumption on $$T$$.
• @Gonzalo $\forall x \phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula. – Henno Brandsma Jan 26 at 11:49