How hard is it to get "absolutely" no amorphous sets? A beautiful and surprising (to me at least) result around the axiom of choice is that, while full $\mathsf{AC}$ is preserved by forcing, a model of $\mathsf{ZF}$ + "There are no amorphous sets" may have a (set-)generic extension in which there do exist amorphous sets. This was proved by Monro, On generic extensions without the axiom of choice; see also Asaf Karagila's summary here.
I'm interested in not-too-strong sufficient conditions on a c.t.m. $\mathcal{M}\models\mathsf{ZF}$ to have no generic extensions in which amorphous sets exist. Specifically, I'm curious if the following model-theoretic condition does the job:

Say that a c.t.m. $\mathcal{M}\models\mathsf{ZF}$ is expansive iff for every first-order theory $T\in \mathcal{M}$ in a finite language with infinite models, every infinite set in $\mathcal{M}$ is the underlying set of a model of $T$ in $\mathcal{M}$.

(Note that such an $\mathcal{M}$ does correctly compute whether such a $T$ has infinite models.) For instance, expansiveness prevents the existence of infinite Dedekind-finite sets, since we can take $T$ to be the theory of a discrete linear order. My question, then, is how expansiveness, amorphousness in particular, and forcing interact:

*

*Is every set-generic extension of an expansive c.t.m. also expansive?


*If not, is there a set-generic extension of an expansive c.t.m. which has amorphous sets?
(Incidentally, it's not immediately clear to me that expansiveness is any stronger than "Infinite = Dedekind-infinite"!)
 A: Turning my comment into an answer, an $X$ which is the universe of any finitely axiomatized theory with an infinite model must be orderable, and there must be a bijection between between $X$ and $X^2.$ Both of these follow from $X$ satisfying a large finite fragment of PA. In fact this is a characterization of such $X$ by the E.M. model construction, treating $X$ equipped with an arbitrary order as the generating set of indiscernibles. The resulting model is equinumerous with $X$ since $|X|=|X|^2$ implies $|X|=|X|^{<\omega}.$
Just from $|X|=|X|^2,$ we can deduce that a model is expansive precisely if it models choice, by Tarski's equivalence of AC with every infinite set being in bijection with its square. Trivially, such models satisfy generic non-existence of amorphous sets.
There are still a lot of open questions regarding generic non-existence of amorphous sets. Asaf's posts summarize the state of knowledge regarding this principle: it's strictly stronger than nonexistence of amorphous sets, since it fails in the Cohen model, and it's implied by "all sets are almost well-orderable." I don't even know how to show the latter is strictly stronger.
