Give counterexamples to $\Th(K_1 \cap K_2) \subseteq \Th(K_1)\cup \Th(K_2)$ and $\Mod(\Gamma \cap \Delta) \subseteq \Mod(\Gamma) \cup \Mod(\Delta)$

In this context, $K_1$ and $K_2$ are classes of $\Sigma$-structures and $\Gamma$ and $\Delta$ are sets of formulas and $$\Th(K) = \{ \varphi \in \operatorname{Sent}_\Sigma \mid \mathcal{A} \models \varphi, \, \forall \mathcal{A} \in K\}$$ and $$\Mod(\Phi) = \{ \mathcal{N} \mid \mathcal{N} \models \phi, \, \forall \phi \in \Phi\}$$ I have already shown that the inclusion "$\supseteq$" always holds and I have been trying to come up with a counterexample showing that the other inclusion, "$\subseteq$" does not hold in general, but failed at the attempt. Any hint on how to proceed would be highly appreciated. Thank you!

UPDATE In a new attempt to come up with something, let $\varphi = \exists x P(x)$, $\psi_1 = \exists x P(x) \rightarrow \exists x R(x)$ and $\psi_2 = \exists xP(x) \rightarrow \exists xQ(x)$. If our universe only has one element $a$, that is $A = \{ a \}$. Let $\Gamma = \{\varphi, \psi_1 \}$ and $\Delta = \{ \varphi, \psi_2 \}$. Then $\Gamma \cap \Delta = \{ \varphi \}$. Then, a model $\mathcal{N}$ for $\Gamma \cap \Delta$ can be constructed by letting $P^{\mathcal{N}}(a) = 1$ and $Q^{\mathcal{N}}(a) = 0$. However, $\mathcal{N} \not \models \psi_1$ and $\mathcal{N} \not \models \psi_2$, therefore, $\mathcal{N} \not \in \Mod(\Gamma) \cup \Mod(\Delta)$. Is this correct?

  • $\begingroup$ "I have already shown that the other inclusion always holds". Which one do you mean, where you write "the other"?? $\endgroup$ Jun 28, 2017 at 19:30
  • 1
    $\begingroup$ @MichaelHardy I meant the inclusion that goes the other way, $\supseteq$. I'll make an edit if it is not clear what I meant. $\endgroup$
    – user313212
    Jun 28, 2017 at 19:44
  • $\begingroup$ Do you mean that BOTH of the inclusions $\supseteq$ hold? $\endgroup$ Jun 28, 2017 at 20:07

2 Answers 2


Let $K_1$ be the class of all abelian groups in which the square of every element is the identity.

Let $K_2$ be the class of all abelian groups in which the cube of every element is the identity.

Then in the theory of $K_1\cap K_2$ the following is true: $\forall x\ \forall y\ x=y.$

But that is not a member of the union of the two theories, since it is a member of neither of them.

  • $\begingroup$ That would be the theory $K_1\cup K_2$. $\endgroup$ Jun 28, 2017 at 21:22
  • $\begingroup$ This is a great example! Thank you very much. I'm accepting your answer, but do you know if the counterexample I wrote on the update to my question is correct? $\endgroup$
    – user313212
    Jun 28, 2017 at 22:32
  • $\begingroup$ @AndrésE.Caicedo : There was a typo: I omitted the word "of". I meant the theory theory OF $K_1\cap K_2.$ The class $K_1\cap K_2$ contains precisely those abelian groups in which the cube of every element is the identity AND the square of every element is the identity, and therefore contains only the group with a single element. That group satisfies some propositions not satisfied by some members of $K_1,$ and likewiess $K_2. \qquad$ $\endgroup$ Jun 28, 2017 at 23:29

$\newcommand{\Mod}{\operatorname{Mod}}\newcommand{\Th}{\operatorname{Th}}$I think the easiest cases to consider would be (a) when $K_1$ and $K_2$ are disjoint sets and (b) when $\Gamma$ and $\Delta$ are disjoint sets. What can you say, in those cases, about $\Th(K_1 \cap K_2)$ and $\Mod(\Gamma \cap \Delta)$?

  • $\begingroup$ I am sorry but since in both cases the intersection is the empty set, doesn't the inclusion hold vacuously? $\endgroup$
    – user313212
    Jun 28, 2017 at 22:30
  • $\begingroup$ I mean, isn't $\Th(K_1 \cap K_2) = \emptyset$ and $\Mod(\Gamma \cap \Delta) = \emptyset$? $\endgroup$
    – user313212
    Jun 28, 2017 at 22:33
  • $\begingroup$ @user313212 Say that $\phi$ is the sentence $\forall x (x = x)$. It sounds like you're saying that this sentence is not in $\Th(\emptyset)$. Looking at the definition of $\Th(K)$, what would it mean for $\phi$ not to be in that set? $\endgroup$ Jun 28, 2017 at 22:57
  • $\begingroup$ You are right, what I wrote above is false. So really $\Th(\emptyset)$ is just the set of all sentences that are tautologies, right? Those that are modeled by the empty set. But any of those sentences would be in both $\Th(K_1)$ and $\Th(K_2)$ as well. $\endgroup$
    – user313212
    Jun 28, 2017 at 23:08
  • $\begingroup$ In your example, $\forall x (x = x)$ is always true and hence $\mathcal{A} \models \phi$ for any model $\mathcal{A} \in K_1, K_2$ $\endgroup$
    – user313212
    Jun 28, 2017 at 23:09

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