Can two first-order theories agree on infinite structures but disagree on finite structures?

Question

This is a dual question to my previous question, here: Can two first-order theories agree on finite structures but disagree on infinite structures?. My current question is, does there exist a first-order signature $L$ and two first-order $L$-theories $T$ and $T'$ such that both $T$ and $T'$ have infinite models and also arbitrarily large finite models, and the infinite models of $T$ and $T'$ are the same, but the finite models of $T$ and $T'$ are not the same? Also, bonus question, if the answer is yes, is it possible for either or even both of $T$ and $T'$ to be finitely axiomatizable?

Answer 1

Converting my comments into an answer:

There are easy finitely axiomatizable examples, e.g. $T = \{(\exists x \exists y\, (x\neq y))\rightarrow P\}$ and $T' = \{P\}$. [Here $P$ is a proposition symbol, i.e., a $0$-ary relation symbol, which is set to either true or false in a structure. If you're not comfortable with proposition symbols, you can replace $P$ with $\forall x\, P(x)$ everywhere in this answer.]

These two sentences have the same infinite models, and have arbitrarily large finite models, but they are not equivalent, since the first has a model of size $1$ in which $P$ is false.

A more interesting variant is the following: Are there theories $T$ and $T'$ which have the same infinite models, both have arbitrarily large finite models, and for all $n\in \omega$, $T$ and $T'$ disagree about models of size $\geq n$?

The answer is no if we additionally require $T$ and $T'$ to be finitely axiomatizable. Suppose for contradiction we have finitely axiomatizable $T$ and $T'$ as above. For all $n\in \omega$, we can pick a finite structure $M_n$ with $|M_n|\geq n$ and either $M_n\models T$ and $M_n\not \models T'$, or $M_n\models T'$ and $M_n\not \models T$. By passing to a subsequence, without loss of generality we can assume that $M_n\models T$ and $M_n\not\models T'$ for all $n\in \omega$. Now let $\mathcal{U}$ be a non-principal ultrafilter on $\omega$, and take the ultraproduct $M = \left(\prod_{n\in \omega} M_n\right){/}\mathcal{U}$. Since both the class of models of $T$ and the class of non-models of $T'$ are elementary, they are closed under ultraproducts, so $M\models T$ and $M\not\models T'$.

On the other hand, the answer is yes in general, and we can even find examples where one of $T$ or $T'$ is finitely axiomatizable. Let $T'$ be the empty theory (which is finitely axiomatizable!). Let $\varphi_n$ be the sentence asserting that there are at most $n$ elements. Let $T$ consistent of the sentences $\varphi_n\to P$ for all $n\in \omega$.

Then $T$ and $T'$ agree on infinite models. Every infinite structure is a model of both $T$ and $T'$, since infinite structures fail to satisfy all $\varphi_n$. And both $T$ and $T'$ have arbitrarily large finite models (any finite structure in which $P$ is true is a model of both). But for all $n$, letting $M_n$ be a structure of size $n$ in which $P$ is false, $M_n\models T'$ but $M_n\not\models T$, since the sentence $\varphi_{n+1}\to P$ is false in $M_n$.