Fix an arbitrary first-order theory $T$ and a monster model $\mathfrak{U}\models T$. Recall that a $\emptyset$-definable subset $D\subseteq\mathfrak{U}$ is said to be "stably embedded" if, for any formula $\varphi(\overline{v},\overline{c})$ with arbitrary parameters, the set $\varphi(D,\overline{c})$ is $D$-definable. We say also that a single-variable formula $\theta(v)$ without parameters is stably embedded if $\theta(\mathfrak{U})\subseteq\mathfrak{U}$ is stably embedded. Now, if $T$ is stable, then every formula is stably embedded. (See e.g. Tent and Ziegler, Corollary 8.3.3.) What can we say in the opposite direction? The question I want to ask is this:
If every formula of $T$ is stably embedded, then is $T$ stable?
However, there are easy counterexamples to this. For instance, if $T=\text{DLO}$, then any two elements of $\mathfrak{U}$ have the same type over $\emptyset$, and hence the only $\emptyset$-definable sets are $\emptyset$ and $\mathfrak{U}$, each of which is clearly stably embedded. But $\text{DLO}$ is not stable. So, certainly there are unstable theories in which every formula is stably embedded. However, I find this example slightly unsatisfying, since the only $\emptyset$-definable subsets are trivial. Unfortunately, things don't get any better if we demand that $T$ has non-trivial $\emptyset$-definable subsets, since (for example) adding a constant symbol to the $\text{DLO}$ example will not change anything. So, I'm not exactly sure what the right formulation of my question is. Very vaguely:
Is there a "natural" property $P$ of $\emptyset$-definable subsets of a theory such that, if $T$ has at least one $\emptyset$-definable subset satisfying property $P$, and every formula of $T$ is stably embedded, then $T$ is stable?
Apologies for the enormous vagueness of this question, and please feel free to close if it is too imprecise. But I would be interested in any insight anyone might have!
Certainly $P$ should be stable and stably embedded. It must also be in some sense "large" for otherwise you can take the disjoint union of a stable theory and an arbitrary theory. One way to make this precise is this: you have a theory $T$ and a distniguished definable set $P$ which is stable and stably embedded such that $M$ and $P(M)$ have the same cardinality for every model $M$. Then does it follow that $T$ is stable? I suspect the answer is yes but I'm not sure.