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In Jech's book Set Theory, the author defines definable over a model:

$$\text{def}_A(M) = \{X \subseteq M : X \text{ is definable over }(M,\epsilon,A \cap M)\},$$

where $A \cap M$ is considered a unary predicate. So let $X$ be a definable set over $(M,\epsilon,A \cap M)$; in other words, there exist a formula $\varphi$ and $a_1, \ldots, a_n \in M$ then there exists a formula $\varphi$ such that $X = \{x \in M : (M,\epsilon,A \cap M) \models \varphi(x,a_1, \ldots, a_n)\}$.

Does this mean that $a_i \in A \cap M$ for $I = 1, \ldots, n$? I am not sure what is the role the unary predicate $A \cap M$ plays in this definition. Any help will be appreciated.

Asaf Karagila
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    It means that all the parameters $a_i$ are in $M$ They may or may not be in $A$. This is just as in ordinary constructibility, the special case of relative constructibility where $A = \emptyset$. Indeed, for any $A\in L$ we have $L[A] = L$. (Why?) – BrianO Oct 14 '21 at 20:42

2 Answers2

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No, the $a_i$s are not required to come from $A\cap M$. Rather, in $\mathsf{def}_A(M)$ the unary predicate naming ($M$'s intersection with) $A$ is available for use in the definition $\varphi$, similarly to how we use individual elements as parameters in definitions.

For example, we trivially have the following:

Suppose $(M;\in)$ is infinite. Then there is an $X\subseteq M$ which is not definable in $\mathcal{M}$; however, for every $X\subseteq M$ there is some $A\subseteq M$ such that $X\in\mathsf{def}_A(\mathcal{M})$.

The first result is just a counting argument; for the second result, take $A=X$ (or indeed $A=X\cup Y$ for any $Y$ disjoint from $M$) and $\varphi(x)\equiv A(x)$ (no $a_i$s are needed here at all).

Noah Schweber
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It's an extra predicate you have in building the formulas. So for instance, $\varphi(x,a)$ could say $x\in a\land A(x)$ which is interpreted in $(M,\in, A\cap M)$ as meaning $x\in a\land x\in A\cap M,$ so e.g. this defines the set $X=(a\cap M)\cap (A\cap M).$

spaceisdarkgreen
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    So, $\text{def}_A(M)$ is just $\text{def}(M)$ with an extra condition, i.e. $A(x)$, we can use when defining $\theta$? –  Oct 14 '21 at 20:55
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    @BobVance Yep, and with a bigger language you can (generally speaking) define more stuff, as Noah points out. – spaceisdarkgreen Oct 14 '21 at 21:01