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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.

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.
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 $\mathcal{A}=L_{\omega_1^{CK}}$.