Suppose $(M, \mathcal X) \models ACA\_0$. Recall that a subset $A \subseteq M$ is $inductive$ over $M$ if $M$ satisfies all instances of induction in the expanded language with a predicate for $A$. Suppose $A$ is inductive and let us denote by $(M, \mathcal X[A])$ the model whose second order part consists of all sets definable from parameters in $\mathcal X$ and $A$. What I want to know is whether or not this is necessarily again a model of $ACA\_0$.


If it helps, my intended application concerns countable models ($M$ and $\mathcal X$ are both countable) and I know that the $A$ I have in mind is in fact inductive in the larger language i.e. $M$ satisfies all instances of induction in the language with predicates for $A$ and all elements of $\mathcal X$ (note that every set in a model of $ACA\_0$ is inductive). 


I suppose more generally I'm interested in knowing, given $A \notin \mathcal X$ which is inductive over $M$, what do I need to require of a $\mathcal Y$ so that $A \in \mathcal Y$, $\mathcal X \subseteq \mathcal Y$ and $(M, \mathcal Y) \models ACA\_0$?



If one adds an inductive subset to a model of $ACA_0$, do we always get a new model of $ACA_0$?
The answer to your (first) question is in the negative.


More explicitly: given any countable nonstandard model $M$ of PA, there are inductive subsets $A$ and $B$ of $M$ such that the expansion $(M,A,B)$ fails to satisfy induction in the extended language. I know two ways to achieve this:


**(1)** Use (Cohen) forcing to construct $A$ and $B$. This was first done by Andrzej Mostowski in his paper: *A remark on models of the Gödel-Bernays axioms for set theory. Sets and classes (on the work by Paul Bernays), pp. 325–340. Studies in Logic and the Foundations of Math., Vol. 84, North-Holland, Amsterdam, 1976.* 


Mostowski's strategy is similar to the proof of the theorem credited to Hugh Woodin in [this MO answer](https://mathoverflow.net/questions/70211/mutually-generics) of Joel Hamkins.


**(2)** Use tools of model theory (Beth's theorem, or that of Svenonius) to show that there are countable nonstandard models of PA (or ZF) that carry distinct (indeed continuum many) *full inductive satisfaction classes*. It is easy to see that if $A$ and $B$ are full satisfaction classes over a model $M$ of PA such that the expansion $(M,A,B)$ satisfies PA in the extended language, then $A=B$, so if $A$ and $B$ are distinct full inductive satisfaction classes over a model $M$ of PA (or ZF), then the expansion $(M,A,B)$ does not satisfy PA (or ZF) in the extended language.


So $A$ and $B$ are as in (1) or (2) above and $\cal{X}$ is chosen as the collection of all subsets of $M$ that are definable in $(M,A)$, then the model $(M,\cal{X}$$[B]$) fails to satisfy $ACA\_0$ even though $(M,\cal{X})$ does satisfy $ACA\_0$ and $B$ is inductive.


As for the second question: the only necessary and sufficient condition for $(M,\cal{X}$$[A]$) to satisfy $ACA\_0$ that I know of is that $(M,X,A)$ satisfies PA in the extended language for every $X \in \cal{X}$.


