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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?

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+1 for questioning yourself. Self-critique is hard to come by these days. – Asaf Karagila Dec 13 '12 at 7:30
I added the $\langle$, $\rangle$ and $<$ to the title: as written I initially interpreted the question to be in the language of ordered which case the result is highly false. – Pete L. Clark Dec 13 '12 at 14:37
up vote 5 down vote accepted

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.

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very clear and concise. Thank you! – Nik Kumar Dec 13 '12 at 4:00
should ∀u1⋯∀un(S(u1,…,un)⟶∃x(x,u1,…,un)) actually read ∀u1⋯∀un(S(u1,…,un)⟶∃xφ(x,u1,…,un)? – Nik Kumar Dec 13 '12 at 5:33
We want $u_i$ in both parts. – André Nicolas Dec 13 '12 at 5:35
I was more preoccupied with whether or not we wanted a phi in the implied statement? – Nik Kumar Dec 13 '12 at 5:38
Sorry for the typo! Yes, there was a missing $\varphi$. I thought I saw an $r_1,\dots,r_n$ in your comment, and fixated on that. But it looks as if they were $u_i$, certainly are now. I find math not set in TeX hard to read! – André Nicolas Dec 13 '12 at 5:45

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 $M=(2{\bf Z},+),N=({\bf Z},+)$).

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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 $\mathbf{A} \preceq \mathbf{B}$

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thanks! I'm a bit lost with how to go about this and would appreciate any guidance – Nik Kumar Dec 13 '12 at 3:42
@NikKumar I've updated with a suggestion for how to solve this problem in particular. – Deven Ware Dec 13 '12 at 3:43
thanks! so now it really just boils down to proving the lemma, then showing that the lemma can be applied to our case of Q and R? – Nik Kumar Dec 13 '12 at 3:46
@NikKumar yep thats right – Deven Ware Dec 13 '12 at 3:49
I haven't yet given proving the lemma much thought, but is it's application to Q and R not trivial? Clearly, for every finite set of Q, there is an automorphism of R that fixes that set, then can map a point b in R to a point in Q? Or am I not understanding this properly? – Nik Kumar Dec 13 '12 at 3:55

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