The $\forall$-theory of $(\mathbb{R}; \leq)$

Question

Consider the structure $(\mathbb{R};\leq)$. What is an axiomatization of its $\forall$-theory? I conjecture that the axioms for linear orders suffice. That is, all you need are reflexivity, anti-symmetry, transitivity, and connectedness. Is this true? Also, bonus question, is the $\forall$-theory of any linear order $(L;\leq)$ with more than one element axiomatized by the axioms for linear orders?

Answer 1

Every countable linear order embeds into $\mathbb{R}$ (or even $\mathbb{Q}$). Consequently, if $\varphi$ is not a consequence of the linear order axioms, then $\varphi$ fails in some substructure of $\mathbb{R}$. Since $\forall^*$-sentences are preserved under taking substructures, we would get that $\varphi$ fails in $\mathbb{R}$ itself. This means that every $\forall^*$-sentence true in $\mathbb{R}$ must already be a consequence of the linear order axioms (applying downward Lowenheim-Skolem appropriately), answering your first question positively.

Your second question meanwhile has an easy negative answer: consider for instance $$\forall x,y,z(x=y\vee y=z\vee x=z),$$ which is true in any linear order with two elements. More generally, any finite linear order satisfies $\forall^*$-sentences not implied by the linear order axioms. However, this is the only sort of obstacle that can occur: the $\forall^*$-theory of an infinite linear order is indeed just the set of $\forall^*$-consequences of the linear order axioms. (To see this, note that any infinite linear order has an elementary extension into which $\mathbb{Q}$ embeds, and then apply the first paragraph of this answer.)