Questions on operations in model theory The intersection of two subgroups is always a subgroup, but the union of two subgroups is generally not a subgroup. Neither is the complement of a subgroup. This holds for other algebraic structures such as rings and linear spaces.
This fact suggests the following holds in model theory. Suppose $T$ is a $\mathcal{L}$-theory, $\phi$ and $\varphi$ are consistent $\mathcal{L}$-sentences in $T$, $M$ and $N$ are models of $T$. If $M\models \phi$ and $N\models\phi$, then $M\cap N$ is a model of $T$ and $M\cap N\models \phi$. However, $M\cup N$ and $N^c$ are (generally) not a structure in $T$.
I wonder if there is such a conclusion in model theory. How is it reflected in model theory?
 A: In the comments and chat on your previous question, Noah Schweber and I both gave you examples showing that in general, the intersection of two (elementary) submodels need not be an (elementary) submodel. TomKern has given you another example in the comments to this question. But I will try one more time.
Let $T$ be the theory of groups. Let $\varphi$ be the sentence $\exists x\, (x\neq e)$. The models of $T\cup \{\varphi\}$  are the non-trivial groups (the groups which contain some non-identity element). Now I will give you two counterexamples:

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*Let $M = (\mathbb{Z},+)$, and let $N = S_3$. Then $M\models \varphi$ and $N\models \varphi$. But since no permutation of the set with three elements is also an integer, $M\cap N = \varnothing$ is not even a group. The point of this example is that you forgot a hypothesis in your question. You want to assume $M$ and $N$ are both submodels of another model.


*Let $U = (\mathbb{Z}/6\mathbb{Z},+)$, the cyclic group of order $6$. Let $M = \langle 2\rangle$ and let $N = \langle 3\rangle$. These are subgroups of $U$ of order $3$ and $2$, respectively. So $M\models \varphi$ and $N\models \varphi$. But $M\cap N = \{0\}$, so $M\cap N\not\models \varphi$.
My example of $U = (\mathbb{R},<)$, $M = (\mathbb{Q},<)$, and $N = (\mathbb{Q}+\pi,<)$ from your previous question shows that the proposition is false even if we require $M$ and $N$ to be elementary substructures of $U$.

Now here's some additional information that you might find helpful.

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*For any language $L$, if $M$ and $N$ are substructures of $U$, then $M\cap N$ is a substructure of $U$. If $\varphi$ is a universal sentence,  i.e. one of the form $\forall \overline{x},\varphi(\overline{x})$, where $\varphi$ is quantifier-free, then $M\models \varphi$ implies $M\cap  N\models \varphi$. This expains why groups,  rings, vector spaces, etc. are closed under intersection: they are axiomatized by universal sentences.


*If $L$ contains no function symbols of  arity $>1$ (i.e. only relation symbols,  constant symbols, and unary function symbols) then if $M$ and $N$ are substructures of $U$, $M\cup N$ is also a substructure of $U$. If  $\varphi$  is an existential sentence,  i.e. one of the form $\exists \overline{x},\varphi(\overline{x})$, where $\varphi$ is quantifier-free, then $M\models \varphi$ implies $M\cup  N\models \varphi$.


*If $L$ contains a function symbol of  arity $>1$, $M\cup N$ is typically  not  even a substructure of  $U$, much less a model of $\varphi$.


*If $L$ is a relational language (no constant symbols or function symbols), then every subset of a structure $U$ is the domain  of a substructure. It follows that if  $M$ is a substructure of $U$, then the  complement  $U\setminus M$ is a substructure of $U$. But there is  no connection whatsoever between sentences true in  $M$ and sentences true in $U\setminus M$.
