# $dcl(A)=\{x \in M : x$ definable from $A \}$, Show $dcl(dcl(A)=A$

Marker exercise 1.4.10c

$dcl(A)=\{x \in M : x$ definable from $A \}$, Show $dcl(dcl(A)=A$

I assume "$x$ definable from $A$" means "$\{x\}$ is $A$-definable."

Definition 1.3.1: Let $\mathcal{M} = (M, . . .)$ be an $\mathcal{L}$-structure. We say that $X \subseteq M^n$ is definable if and only if there is an $\mathcal{L}$-formula $\phi(v_1 , . . . , v_n , w_1 , . . . , w_m )$ and $b \in M^m$ such that $X = \{a \in M^n : M \models \phi(a, b)\}$. We say that $\phi(v, b)$ defines $X$. We say that $X$ is $A$-definable or definable over $A$ if there is a formula $\phi(v, w_1 , . . . , w_l )$ and $b \in A^l$ such that $\phi(v, b)$ defines $X$.

We can define every $\{a\} \subseteq A$ by a formula $\phi(x,a) \mathrel{\mathop:}=x=a$ since $a \in A$. Plus there are possibly other $\emptyset$-definable singletons depending on our language which are $A$-definable. If we have a language of abelian groups $\mathcal{L}=(0,+)$ and although $0 \notin A$, we can define $0$ by the following formula: $\phi(x,a) \mathrel{\mathop:}=x=0$.

So for the example I have given $dcl(A)$ would be a superset of $A$, $0 \in dcl(A)$, consequently $0 \in dcl(dcl(A))$.

What is wrong with my reasoning?

• There is a typo in the exercise. It should read $\mathrm{dcl}(\mathrm{dcl}(A))=\mathrm{dcl}(A)$. – Andrés E. Caicedo Aug 4 '13 at 23:34

As Andres Caicedo noted, there is a typo in the exercise. It should read $dcl(dcl(A))=dcl(A)$.