Let $L$ be a first-order language. A class $K$ of $L$-structures is said to be infinitely axiomatizable if it is axiomatizable but not finitely axiomatizable. My first question is, does there exist a language $L$ for which there are two classes $K$ and $K'$ which are both infinitely axiomatizable, but whose union $K \cup K'$ is finitely axiomatizable? My second question is, does there exist a language $L$ for which there are two disjoint classes $K$ and $K'$ such that $K$ is infinitely axiomatizable, $K'$ is finitely axiomatizable, and the union $K \cup K'$ is finitely axiomatizable? I am requiring that $K$ and $K'$ are disjoint in the second question to make the question more interesting and rule out trivial examples, like when $K'$ is the class of all $L$-structures.
1 Answers
I think it's helpful to think about problems like this from a topological perspective.
Ignoring set/class issues, consider the topology on the class $\mathbb{S}_L$ of all $L$-structures generated by the base $$\{\{\mathcal{M}: \mathcal{M}\models\varphi\}:\varphi\in Sent_L\}.$$ Trivially we have "axiomatizable $\iff$ closed." Much more significantly, by the compactness theorem we have "finitely axiomatizable $\iff$ clopen," and in fact this is just a restatement of the compactness theorem. (Note that if we were to do this with a non-compact logic in place of $\mathsf{FOL}$ all we would be guaranteed is "finitely axiomatizable $\color{red}{\implies}$ clopen.")
The fact that we can topologically characterize finite axiomatizability immediately gives a strong negative answer to your second question on general topological grounds: if I partition a clopen set into two open pieces, each of those pieces is clopen. However, the first question is left untouched by this reasoning and in fact ultimately has a positive answer:
Consider the language $L=\{A,B\}$ consisting of two unary relations. Let $K$ be the class of $L$-structures in which there are at least as many $A$s as $B$s, and let $K'$ be the class of $L$-structures in which there are at least as many $B$s as $A$s. We have $$K\cup K'=\mathbb{S}_L,$$ but it's easy to show that both $K$ and $K'$ are infinitely axiomatizable.
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1The topological point of view is also useful to understand your counter-example: the space is $(ℕ∪{∞})^2$ with the obvious topology, and we can draw the two closed-and-not-open subspaces whose union is everything (their intersection is ${(∞,∞)}$). There is even a counter-example in $ℕ∪{∞}$: the even numbers and the odd numbers. This translates as: the signature is empty, $K = {\text{sets of even size}}$ and $K' = {\text{sets of odd size}}$. – Dabouliplop Jan 01 '22 at 21:04
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1(Err, the intersection is not ${(∞,∞)}$ but the diagonal. I was thinking about another case.) – Dabouliplop Jan 01 '22 at 23:43