The 8 combinations of finitely axiomatizable for full, universal, and existential theories. Let $L$ be a signature (in the sense of model theory), and let $T$ be an $L$-theory. Also, let $T_\forall$ (respectively, $T_\exists$) be the deductive closure of the universal (respectively, existential) sentences of $T$. Prima facie, there are $8$ possible combinations of finite/non-finite axiomatizability for each of $T$, $T_\forall$, and $T_\exists$. For instance, they could all be finitely axiomatizable, or they could all be non-finitely axiomatizable, or $T_\forall$ could be finitely axiomatizable while both $T$ and $T_\exists$ are non-finitely axiomatizable, etc. My question is, which of these $8$ combinations are actually realized? In other words, for each of those $8$ combinations, I want either an example of a signature $L$ and a $L$-theory $T$ which is an example of that combination along with the proof that it is indeed an example, or, a proof that there is no signature $L$ and no $L$-theory $T$ which has that combination.
 A: Here are examples of all $8$ of the possible behaviors. I tried my best to give the simplest most "combinatorial" and transparent examples I could think of. If anyone has suggestions of simpler examples, I'd be happy to hear them. Also, I thought this through and wrote it up fairly quickly, so it's quite possible there are mistakes in some of these examples. Let me know if you find any!
Note that any consistent existential sentence has a finite model. So if $T_\exists$ has no finite model, $T_\exists$ is not finitely axiomatizable. We'll use this observation a few times.

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*$T$, $T_\forall$, $T_\exists$ all FA: The empty theory.


*$T$, $T_\forall$, $T_\exists$ all $\lnot$FA: We can take any non-finitely-axiomatizable theory axiomatized by quantifier-free sentences, since then $T = T_\forall = T_\exists$. For example, in the language with a constant symbol $c$ and a unary function symbol $f$, $\{f^n(c)\neq c\mid n\in \omega\}$.


*$T$ and $T_\exists$ $\lnot$FA, $T_\forall$ FA:  The theory of an infinite set (in the empty language). This is an existential theory, so $T = T_\exists$ is $\lnot$FA (since it has no finite models), and $T$ has no non-trivial universal consequences.


*$T$ and $T_\forall$ $\lnot$FA, $T_\exists$ FA: Similarly to the last point, it suffices to find a $\lnot$FA universal theory with no non-trivial existential consequences. For example, in the language with a binary relation $R$, consider the theory which says "there is no $R$-cycle of length $n$" for all $n\in\omega$.


*$T$ and $T_\forall$ FA, $T_\exists$ $\lnot$FA: The theory of non-empty dense linear orders without endpoints. Here $T_\forall$ is the theory of dense linear orders without endpoints, which is FA. $T_\exists$ says that every finite linear order embeds in a model, and this is $\lnot$FA because it has no finite models.


*$T$ $\lnot$FA, $T_\forall$ and $T_\exists$ FA: Consider the language with a unary predicate $P$. Let $T$ have one axiom for each $n\in \omega$, which says $(\exists x\, P(x))\rightarrow (\exists^{\geq n} x\, \top)$. A model of $T$ is either a set in which no elements satisfy $P$ or an infinite set in which $P$ is interpreted arbitrarily. $T$ is not finitely axiomatizable, but it has no non-trivial universal or existential consequences.


*$T$ and $T_\exists$ FA, $T_\forall$ $\lnot$FA: Consider the theory in the language with a unary relation $G$, a binary relation $R$, and a ternary relation $E$. The theory says: (1) If $R(a,b)$, then $G(a)$ and $G(b)$. (2) If $E(a,b,c)$, then $\lnot G(a)$, $G(b)$, and $G(c)$. (3) $R$ is a graph relation on $G$ (it is symmetric and antireflexive). (4) For all $a$ such that $\lnot G(a)$, $E(a,y,z)$ is an equivalence relation on $G$ with two classes, establishing a bipartition of the graph $(G,R)$. That is, for all $b$ and $c$, if $R(b,c)$, then $\lnot E(a,b,c)$. (5) There exists $a$ such that $\lnot G(a)$. I have given a finite axiomatization of $T$, and $T_\exists$ just says that there exists an element $x$ satisying $\lnot G(x)$, $\lnot R(x,x)$ and $\lnot E(x,x,x)$, so this is also finitely axiomatizable. But $T_\forall$ says, in addition to (1)-(4), that the graph $R$ has no $n$-cycles for all odd $n\geq 3$. The point is that the existence of a bipartition implies these conditions, but since $T_\forall$ can't ensure the existence of an element $a$ satisfying $\lnot G(a)$, so that $E(a,x,y)$ witnesses that the graph is bipartite, $T_\forall$ has to rule out odd cycles explicitly.


*$T$ FA, $T_\forall$ and $T_\exists$ $\lnot$FA: The idea is to combine examples 5 and 7. Take the previous example (7), add a new binary relation symbol $<$, and add universal axioms asserting that $<$ is a dense linear order without endpoints on the set of $x$ satisfying $\lnot G(x)$. The same considerations as in 7 show that $T_\forall$ is not finitely axiomatizable. And now $T_\exists$ is not finitely axiomatizable, because it says that every finite linear order embeds in a model.
