# Question on usage of Löwenheim–Skolem in Beginner Logic

I'm taking a beginner course in mathematical logic, and I was struggling to understand some applications of the downward Löwenheim–Skolem theorem. The version of Löwenheim–Skolem theorem that was presented in the lectures is:

Suppose $$L$$ is countable first order language, and $$\mathcal{A}$$ is an $$L$$-structure. Let $$T$$ be the theory of $$\mathcal{A}$$, i.e. $$T = Th(\mathcal{A}) =\{ \theta: \mathcal{A} \models \theta \text{ where } \theta \text{ is an L-sentence}\}$$. Then, there is a countable model $$\mathcal{B}$$ such that $$\mathcal{B}\models T$$.

$$\Delta$$ was defined to be the sentences which describe a dense linear order with no end points, i.e. $$\Delta = \{ \varphi_1, \dots , \varphi_6\}$$ where for example $$\varphi_1 = \forall x_1, \exists x_2, \ x_2 < x_1$$. The proof of the "Los-Vaught test" was given.

(Los-Vaught Test) For every $$L$$-sentence $$\theta$$, $$\Delta \vdash \theta$$ or $$\Delta \vdash \lnot \theta$$.

The proof starts: Suppose not. Then, $$\Delta \cup \{\theta\}$$ and $$\Delta \cup \{ \lnot\theta\}$$ are both consistent. Godel's completeness theorem tells us (the model existence lemma tells us?) that if it is consistent, then it should have a model. Then Löwenheim–Skolem theorem tells us, that we should have a countable model.

I don't understand the last sentence, namely, why the Löwenheim–Skolem theorem tells us that. How can I know that $$\Delta \cup \{\theta\}$$ and $$\Delta \cup \{ \lnot\theta\}$$ are theories of some $$L$$-structure?

• I just noticed your other question (math.stackexchange.com/q/4054463/7062). I hope you can see that these two questions (and their answers) are essentially the same! Mar 9, 2021 at 3:56

If $$M$$ is a model of $$\Delta\cup\{\theta\}$$, then $$\Delta\cup\{\theta\}\subseteq\mathrm{Th}(M)$$. So any model of $$\mathrm{Th}(M)$$ is a model of $$\Delta\cup\{\theta\}$$.
• Ah, so am I correct to understand that $\Delta \cup \{\theta\}$ has a model $M$ by Model existence lemma, and $Th(M)$ has a countable model $\mathcal{B}$ due to Löwenheim–Skolem. As $\Delta \cup \{\theta\} \subset Th(M)$, we conclude $\mathcal{B} \models \Delta \cup \{\theta\}$? Mar 9, 2021 at 3:50