4
$\begingroup$

Let $\Sigma_1$ and $\Sigma_2$ be sets of $L$-sentences such that no symbol of $L$ occurs in both $\Sigma_1$ and $\Sigma_2$. Suppose $\Sigma_1$ and $\Sigma_2$ have infinite models. Then $\Sigma_1 \cup \Sigma_2$ has a model.

We have some facts about back and forth systems and Vaught's test, and Löwenheim-Skolem's theorem, and by that last one I can say that $\Sigma_1, \Sigma_2$ have models of cardinality equal to the cardinality of $L$, but I don't know where to go from there. For instance, we don't necessarily want to show that $\Sigma_1 \cup \Sigma_2$ is complete, so things like Vaught's test seem irrelevant.

Does this just follow from a slick argument about compactness?

  • 1
    My point is that while the choice in your course is by all means a common and respectable one, it probably cannot be said to be _the_ standard one.2011-10-24

2 Answers 2

5

Because the theories have infinite models, you can find a $\kappa$ large enough that both theories have models of size $\kappa$. What happens if you take two such models, one from each theory, and simply identify the domains with each other?

5

Let $M_1$ and $M_2$ be a models of $\Sigma_1$ and $\Sigma_2$. By Skolem-Löwenheim we can assume without loss of generality that $M_1$ and $M_2$ have the same cardinality. And then, again without loss of generality, we can assume that they have the same set of individuals and just come with different interpretations of their respective parts of the language ...