Let $\mathcal{A}$ be an admissible set, or for simplicity a model of ZF without the power set axiom.
 Let $X \subset \mathcal{A}$. Is there an easy way to show $X$ is $\Sigma$ on $\mathcal{A}$?


PS: $X$ is $\Sigma$ on $\mathcal{A}$ if it is definable in $\mathcal{A}$ by a $\Sigma$ formula i.e.
 a formula built from atomic formulas and their negations using only $\wedge,\vee, \forall x \in y, \exists x$. To be specific assume $\mathcal{A}=L\_{\omega\_1^{CK}}$. Also, I need to consider the specific case of $X$ being a set of (codes of) $\mathcal{L}(\omega\_1,\omega)$-sentences or to be precise: $X$ is a set of $\mathcal{L}$-sentences + infinitary sentences specifying that certain quotient sets are finite. Here $\mathcal{L}$ is some first order language.



How to know that a set is \Sigma?
A frequently useful way to show that a function $F$ is $\Sigma$ definable is to give a definition of it by recursion, in the form $F(x)=G(x,F\upharpoonright H(x))$ where $G$ and $H$ are already known to be $\Sigma$-definable and $H(x)$ is the set of predecessors of $x$ in a well-founded relation $\prec$ (so that the recursion is well-defined). 


It is important here that the function $H$, not just the (morally equivalent) relation $\prec$, be $\Sigma$-definable. 


Although I wrote here about the special case of $\Sigma$-defining *functions*, such definitions often play a key role in $\Sigma$-definitions of other sorts of sets.


