$\langle \Bbb Q,<\rangle$ is an elementary submodel of $ \langle \Bbb R,< \rangle$

Question

I am trying to show that $\langle\Bbb Q,<\rangle$ is an elementary submodel of $\langle\Bbb R,<\rangle$.

I first believed that this problem is quite trivial $-$ I thought all I needed to do was show that $\langle\Bbb Q,<\rangle$ and $\langle\Bbb R,<\rangle$ are elementarily equivalent (which follows since both are dense linear orders without endpoint) and then say that $\Bbb Q$ is obviously contained in $\Bbb R$. However, I'm now questioning myself after examining definitions more closely, in particular those related to this post.

In other words, is it not enough to show that the two models are elementarily equivalent, and one a submodel of the other when trying to show one is an elementary submodel of the other?

Answer 1

Let $\varphi(x,u_1,\dots,u_n)$ be a formula of the language, and suppose that $\exists x\varphi(x,r_1,\dots,r_n)$ is true in the rationals, for specific rationals $r_1,\dots,r_n$. Then by the completeness of the theory, if we let $S$ be the finite collection of order relationships among the $r_i$, then $$\forall u_1\cdots \forall u_n\left(S(u_1,\dots,u_n)\longrightarrow \exists x\varphi(x,u_1,\dots,u_n)\right)$$ is true in the rationals, and hence in the reals. Thus $\exists x\varphi(x,r_1,\dots,r_n)$ is true in the reals. The rest follows from Vaught's Test.

Answer 2

No it is not enough to show elementary equivalence and that one is a subset of the other.

What you need to show is that $$\mathbf{Q} \models \sigma \iff \mathbf{R} \models \sigma$$ for any $\mathcal{L}_{\mathbb{Q}}$-sentence $\sigma$. (where $\mathcal{L}_{\mathbb{Q}}$ is the extension of $\mathcal{L}$ with names for all elements of $\mathbb{Q}$ added in)

EDIT: For your problem of showing $\mathbf{Q} \preceq \mathbf{R}$ I would recommend first showing the following lemma:

If $\mathbf{A} \subseteq \mathbf{B}$ and for every finite subset $K \subseteq A$ and $b\in B$ there is an automorphism $f$ of $\mathbf{B}$ which fixes $K$ and $f(b) \in A$ then $\mathbf{A} \preceq \mathbf{B}$

Answer 3

Another way to show the fact is to notice that the theory of $\bf Q$, the theory of dense linear orderings without endpoints, is satisfied by $\bf R$ and eliminates quantifiers.

If a theory $T$ eliminates quantifiers, then it is true that if we have two structures $M\subseteq N$ both satisfying $T$, then $M\preceq N$ (that follows from the fact that quantifier-free formulas are absolute, which is an easy exercise).

Therefore, whenever you have a model $M$ such that $\operatorname{Th}(M)$ eliminates quantifiers, and $M$ is a submodel of $N$, then to show that it is elementary it is enough to show that they're elementarily equivalent.

In general, $M\subseteq N\wedge M\equiv N$ is strictly weaker than $M\preceq N$ (as I've mentioned in the other post with the example $M=(2{\bf Z},+),N=({\bf Z},+)$).